Solving $int sqrt{x}left(1+sqrt{x}right)^3mathrm{d}x $












0












$begingroup$


$$ int sqrt{x}left(1+sqrt{x}right)^3dx $$
I tried to solve this by using substitution. However, I could not reach the answer.
I tried to replace $1+sqrt{x} = u$ and
$sqrt{x} = u$
But still did not get the answer.I don't want to open parenthesis and solve integral in that way.






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean $$int sqrt{x} cdot left(1+sqrt{x}right)^3;?$$
    $endgroup$
    – Chinnapparaj R
    Dec 17 '18 at 15:35










  • $begingroup$
    Yes,thanks for correcting
    $endgroup$
    – Arif Rustamov
    Dec 17 '18 at 15:40






  • 3




    $begingroup$
    Mutliply out the cube to get four terms. Multiply through by the square root. Then you have an easy sum of powers.
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 15:40






  • 3




    $begingroup$
    Why exactly would you not want to expand the integrand?
    $endgroup$
    – KM101
    Dec 17 '18 at 15:42






  • 1




    $begingroup$
    If you do a substitution you will still have parentheses that need opening: you may get three rather than four terms, but at the cost of the substitution, and for an indefinite integral the cost of re-substitution
    $endgroup$
    – Henry
    Dec 17 '18 at 15:55


















0












$begingroup$


$$ int sqrt{x}left(1+sqrt{x}right)^3dx $$
I tried to solve this by using substitution. However, I could not reach the answer.
I tried to replace $1+sqrt{x} = u$ and
$sqrt{x} = u$
But still did not get the answer.I don't want to open parenthesis and solve integral in that way.






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean $$int sqrt{x} cdot left(1+sqrt{x}right)^3;?$$
    $endgroup$
    – Chinnapparaj R
    Dec 17 '18 at 15:35










  • $begingroup$
    Yes,thanks for correcting
    $endgroup$
    – Arif Rustamov
    Dec 17 '18 at 15:40






  • 3




    $begingroup$
    Mutliply out the cube to get four terms. Multiply through by the square root. Then you have an easy sum of powers.
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 15:40






  • 3




    $begingroup$
    Why exactly would you not want to expand the integrand?
    $endgroup$
    – KM101
    Dec 17 '18 at 15:42






  • 1




    $begingroup$
    If you do a substitution you will still have parentheses that need opening: you may get three rather than four terms, but at the cost of the substitution, and for an indefinite integral the cost of re-substitution
    $endgroup$
    – Henry
    Dec 17 '18 at 15:55
















0












0








0





$begingroup$


$$ int sqrt{x}left(1+sqrt{x}right)^3dx $$
I tried to solve this by using substitution. However, I could not reach the answer.
I tried to replace $1+sqrt{x} = u$ and
$sqrt{x} = u$
But still did not get the answer.I don't want to open parenthesis and solve integral in that way.






calculus integration







share|cite|improve this question











$endgroup$




$$ int sqrt{x}left(1+sqrt{x}right)^3dx $$
I tried to solve this by using substitution. However, I could not reach the answer.
I tried to replace $1+sqrt{x} = u$ and
$sqrt{x} = u$
But still did not get the answer.I don't want to open parenthesis and solve integral in that way.






calculus integration




calculus integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 15:54









Rebellos

15.3k31250




15.3k31250










asked Dec 17 '18 at 15:33









Arif RustamovArif Rustamov

387




387












  • $begingroup$
    Do you mean $$int sqrt{x} cdot left(1+sqrt{x}right)^3;?$$
    $endgroup$
    – Chinnapparaj R
    Dec 17 '18 at 15:35










  • $begingroup$
    Yes,thanks for correcting
    $endgroup$
    – Arif Rustamov
    Dec 17 '18 at 15:40






  • 3




    $begingroup$
    Mutliply out the cube to get four terms. Multiply through by the square root. Then you have an easy sum of powers.
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 15:40






  • 3




    $begingroup$
    Why exactly would you not want to expand the integrand?
    $endgroup$
    – KM101
    Dec 17 '18 at 15:42






  • 1




    $begingroup$
    If you do a substitution you will still have parentheses that need opening: you may get three rather than four terms, but at the cost of the substitution, and for an indefinite integral the cost of re-substitution
    $endgroup$
    – Henry
    Dec 17 '18 at 15:55




















  • $begingroup$
    Do you mean $$int sqrt{x} cdot left(1+sqrt{x}right)^3;?$$
    $endgroup$
    – Chinnapparaj R
    Dec 17 '18 at 15:35










  • $begingroup$
    Yes,thanks for correcting
    $endgroup$
    – Arif Rustamov
    Dec 17 '18 at 15:40






  • 3




    $begingroup$
    Mutliply out the cube to get four terms. Multiply through by the square root. Then you have an easy sum of powers.
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 15:40






  • 3




    $begingroup$
    Why exactly would you not want to expand the integrand?
    $endgroup$
    – KM101
    Dec 17 '18 at 15:42






  • 1




    $begingroup$
    If you do a substitution you will still have parentheses that need opening: you may get three rather than four terms, but at the cost of the substitution, and for an indefinite integral the cost of re-substitution
    $endgroup$
    – Henry
    Dec 17 '18 at 15:55


















$begingroup$
Do you mean $$int sqrt{x} cdot left(1+sqrt{x}right)^3;?$$
$endgroup$
– Chinnapparaj R
Dec 17 '18 at 15:35




$begingroup$
Do you mean $$int sqrt{x} cdot left(1+sqrt{x}right)^3;?$$
$endgroup$
– Chinnapparaj R
Dec 17 '18 at 15:35












$begingroup$
Yes,thanks for correcting
$endgroup$
– Arif Rustamov
Dec 17 '18 at 15:40




$begingroup$
Yes,thanks for correcting
$endgroup$
– Arif Rustamov
Dec 17 '18 at 15:40




3




3




$begingroup$
Mutliply out the cube to get four terms. Multiply through by the square root. Then you have an easy sum of powers.
$endgroup$
– Ethan Bolker
Dec 17 '18 at 15:40




$begingroup$
Mutliply out the cube to get four terms. Multiply through by the square root. Then you have an easy sum of powers.
$endgroup$
– Ethan Bolker
Dec 17 '18 at 15:40




3




3




$begingroup$
Why exactly would you not want to expand the integrand?
$endgroup$
– KM101
Dec 17 '18 at 15:42




$begingroup$
Why exactly would you not want to expand the integrand?
$endgroup$
– KM101
Dec 17 '18 at 15:42




1




1




$begingroup$
If you do a substitution you will still have parentheses that need opening: you may get three rather than four terms, but at the cost of the substitution, and for an indefinite integral the cost of re-substitution
$endgroup$
– Henry
Dec 17 '18 at 15:55






$begingroup$
If you do a substitution you will still have parentheses that need opening: you may get three rather than four terms, but at the cost of the substitution, and for an indefinite integral the cost of re-substitution
$endgroup$
– Henry
Dec 17 '18 at 15:55












2 Answers
2






active

oldest

votes


















4












$begingroup$

Let $u = sqrt{x}+1 implies mathrm{d}x = 2sqrt{x}mathrm{d}u$. It will then be $x = (u-1)^2$. Thus, the integral becomes :



$$int (u-1)^2u^3mathrm{d}u ={displaystyleint}u^5,mathrm{d}u-class{steps-node}{cssId{steps-node-2}{2}}{displaystyleint}u^4,mathrm{d}u+{displaystyleint}u^3,mathrm{d}u =dfrac{u^6}{6}-dfrac{2u^5}{5}+dfrac{u^4}{4} + C $$



Now, substitute for $u= sqrt{x} + 1$ and you should get for the initial integral :



begin{align*}
int sqrt{x}(1+sqrt{x}) mathrm{d}x &=dfrac{left(sqrt{x}+1right)^6}{3}-dfrac{4left(sqrt{x}+1right)^5}{5}+dfrac{left(sqrt{x}+1right)^4}{2} + C\
&= boxed{dfrac{left(sqrt{x}+1right)^4left(10x-4sqrt{x}+1right)}{30}+C}.
end{align*}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can we try $ x^{frac{-1}{2}}+1=u $ by taking $sqrt{x}$ common
    $endgroup$
    – Lakshya Sinha
    Dec 17 '18 at 16:05










  • $begingroup$
    @LakshyaSinha What?
    $endgroup$
    – Rebellos
    Dec 17 '18 at 16:05










  • $begingroup$
    Take $ sqrt{x} $ common from cubic term and suppose that as u
    $endgroup$
    – Lakshya Sinha
    Dec 17 '18 at 16:07










  • $begingroup$
    The person who disliked would be kind enough to point out ? The solution is perfectly clear and correct.
    $endgroup$
    – Rebellos
    Dec 18 '18 at 18:42



















0












$begingroup$

By parts $u=(1+sqrt{x})^3$ and $v'=sqrt{x}$:



$$int sqrt{x}left(1+sqrt{x}right)^3dx=frac{2}{3}x^{frac{3}{2}}left(1+sqrt{x}right)^3-int :xleft(1+sqrt{x}right)^2dx$$



And then by expanding:
$$int :xleft(1+sqrt{x}right)^2dx=frac{x^2}{2}+frac{4}{5}x^{frac{5}{2}}+frac{x^3}{3}$$



So
$$int sqrt{x}left(1+sqrt{x}right)^3dx=frac{2}{3}x^{frac{3}{2}}left(1+sqrt{x}right)^3-frac{x^2}{2}-frac{4}{5}x^{frac{5}{2}}-frac{x^3}{3}+C.$$






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044088%2fsolving-int-sqrtx-left1-sqrtx-right3-mathrmdx%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Let $u = sqrt{x}+1 implies mathrm{d}x = 2sqrt{x}mathrm{d}u$. It will then be $x = (u-1)^2$. Thus, the integral becomes :



    $$int (u-1)^2u^3mathrm{d}u ={displaystyleint}u^5,mathrm{d}u-class{steps-node}{cssId{steps-node-2}{2}}{displaystyleint}u^4,mathrm{d}u+{displaystyleint}u^3,mathrm{d}u =dfrac{u^6}{6}-dfrac{2u^5}{5}+dfrac{u^4}{4} + C $$



    Now, substitute for $u= sqrt{x} + 1$ and you should get for the initial integral :



    begin{align*}
    int sqrt{x}(1+sqrt{x}) mathrm{d}x &=dfrac{left(sqrt{x}+1right)^6}{3}-dfrac{4left(sqrt{x}+1right)^5}{5}+dfrac{left(sqrt{x}+1right)^4}{2} + C\
    &= boxed{dfrac{left(sqrt{x}+1right)^4left(10x-4sqrt{x}+1right)}{30}+C}.
    end{align*}






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Can we try $ x^{frac{-1}{2}}+1=u $ by taking $sqrt{x}$ common
      $endgroup$
      – Lakshya Sinha
      Dec 17 '18 at 16:05










    • $begingroup$
      @LakshyaSinha What?
      $endgroup$
      – Rebellos
      Dec 17 '18 at 16:05










    • $begingroup$
      Take $ sqrt{x} $ common from cubic term and suppose that as u
      $endgroup$
      – Lakshya Sinha
      Dec 17 '18 at 16:07










    • $begingroup$
      The person who disliked would be kind enough to point out ? The solution is perfectly clear and correct.
      $endgroup$
      – Rebellos
      Dec 18 '18 at 18:42
















    4












    $begingroup$

    Let $u = sqrt{x}+1 implies mathrm{d}x = 2sqrt{x}mathrm{d}u$. It will then be $x = (u-1)^2$. Thus, the integral becomes :



    $$int (u-1)^2u^3mathrm{d}u ={displaystyleint}u^5,mathrm{d}u-class{steps-node}{cssId{steps-node-2}{2}}{displaystyleint}u^4,mathrm{d}u+{displaystyleint}u^3,mathrm{d}u =dfrac{u^6}{6}-dfrac{2u^5}{5}+dfrac{u^4}{4} + C $$



    Now, substitute for $u= sqrt{x} + 1$ and you should get for the initial integral :



    begin{align*}
    int sqrt{x}(1+sqrt{x}) mathrm{d}x &=dfrac{left(sqrt{x}+1right)^6}{3}-dfrac{4left(sqrt{x}+1right)^5}{5}+dfrac{left(sqrt{x}+1right)^4}{2} + C\
    &= boxed{dfrac{left(sqrt{x}+1right)^4left(10x-4sqrt{x}+1right)}{30}+C}.
    end{align*}






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Can we try $ x^{frac{-1}{2}}+1=u $ by taking $sqrt{x}$ common
      $endgroup$
      – Lakshya Sinha
      Dec 17 '18 at 16:05










    • $begingroup$
      @LakshyaSinha What?
      $endgroup$
      – Rebellos
      Dec 17 '18 at 16:05










    • $begingroup$
      Take $ sqrt{x} $ common from cubic term and suppose that as u
      $endgroup$
      – Lakshya Sinha
      Dec 17 '18 at 16:07










    • $begingroup$
      The person who disliked would be kind enough to point out ? The solution is perfectly clear and correct.
      $endgroup$
      – Rebellos
      Dec 18 '18 at 18:42














    4












    4








    4





    $begingroup$

    Let $u = sqrt{x}+1 implies mathrm{d}x = 2sqrt{x}mathrm{d}u$. It will then be $x = (u-1)^2$. Thus, the integral becomes :



    $$int (u-1)^2u^3mathrm{d}u ={displaystyleint}u^5,mathrm{d}u-class{steps-node}{cssId{steps-node-2}{2}}{displaystyleint}u^4,mathrm{d}u+{displaystyleint}u^3,mathrm{d}u =dfrac{u^6}{6}-dfrac{2u^5}{5}+dfrac{u^4}{4} + C $$



    Now, substitute for $u= sqrt{x} + 1$ and you should get for the initial integral :



    begin{align*}
    int sqrt{x}(1+sqrt{x}) mathrm{d}x &=dfrac{left(sqrt{x}+1right)^6}{3}-dfrac{4left(sqrt{x}+1right)^5}{5}+dfrac{left(sqrt{x}+1right)^4}{2} + C\
    &= boxed{dfrac{left(sqrt{x}+1right)^4left(10x-4sqrt{x}+1right)}{30}+C}.
    end{align*}






    share|cite|improve this answer











    $endgroup$



    Let $u = sqrt{x}+1 implies mathrm{d}x = 2sqrt{x}mathrm{d}u$. It will then be $x = (u-1)^2$. Thus, the integral becomes :



    $$int (u-1)^2u^3mathrm{d}u ={displaystyleint}u^5,mathrm{d}u-class{steps-node}{cssId{steps-node-2}{2}}{displaystyleint}u^4,mathrm{d}u+{displaystyleint}u^3,mathrm{d}u =dfrac{u^6}{6}-dfrac{2u^5}{5}+dfrac{u^4}{4} + C $$



    Now, substitute for $u= sqrt{x} + 1$ and you should get for the initial integral :



    begin{align*}
    int sqrt{x}(1+sqrt{x}) mathrm{d}x &=dfrac{left(sqrt{x}+1right)^6}{3}-dfrac{4left(sqrt{x}+1right)^5}{5}+dfrac{left(sqrt{x}+1right)^4}{2} + C\
    &= boxed{dfrac{left(sqrt{x}+1right)^4left(10x-4sqrt{x}+1right)}{30}+C}.
    end{align*}







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 17 '18 at 15:54









    Christoph

    12.5k1642




    12.5k1642










    answered Dec 17 '18 at 15:43









    RebellosRebellos

    15.3k31250




    15.3k31250












    • $begingroup$
      Can we try $ x^{frac{-1}{2}}+1=u $ by taking $sqrt{x}$ common
      $endgroup$
      – Lakshya Sinha
      Dec 17 '18 at 16:05










    • $begingroup$
      @LakshyaSinha What?
      $endgroup$
      – Rebellos
      Dec 17 '18 at 16:05










    • $begingroup$
      Take $ sqrt{x} $ common from cubic term and suppose that as u
      $endgroup$
      – Lakshya Sinha
      Dec 17 '18 at 16:07










    • $begingroup$
      The person who disliked would be kind enough to point out ? The solution is perfectly clear and correct.
      $endgroup$
      – Rebellos
      Dec 18 '18 at 18:42


















    • $begingroup$
      Can we try $ x^{frac{-1}{2}}+1=u $ by taking $sqrt{x}$ common
      $endgroup$
      – Lakshya Sinha
      Dec 17 '18 at 16:05










    • $begingroup$
      @LakshyaSinha What?
      $endgroup$
      – Rebellos
      Dec 17 '18 at 16:05










    • $begingroup$
      Take $ sqrt{x} $ common from cubic term and suppose that as u
      $endgroup$
      – Lakshya Sinha
      Dec 17 '18 at 16:07










    • $begingroup$
      The person who disliked would be kind enough to point out ? The solution is perfectly clear and correct.
      $endgroup$
      – Rebellos
      Dec 18 '18 at 18:42
















    $begingroup$
    Can we try $ x^{frac{-1}{2}}+1=u $ by taking $sqrt{x}$ common
    $endgroup$
    – Lakshya Sinha
    Dec 17 '18 at 16:05




    $begingroup$
    Can we try $ x^{frac{-1}{2}}+1=u $ by taking $sqrt{x}$ common
    $endgroup$
    – Lakshya Sinha
    Dec 17 '18 at 16:05












    $begingroup$
    @LakshyaSinha What?
    $endgroup$
    – Rebellos
    Dec 17 '18 at 16:05




    $begingroup$
    @LakshyaSinha What?
    $endgroup$
    – Rebellos
    Dec 17 '18 at 16:05












    $begingroup$
    Take $ sqrt{x} $ common from cubic term and suppose that as u
    $endgroup$
    – Lakshya Sinha
    Dec 17 '18 at 16:07




    $begingroup$
    Take $ sqrt{x} $ common from cubic term and suppose that as u
    $endgroup$
    – Lakshya Sinha
    Dec 17 '18 at 16:07












    $begingroup$
    The person who disliked would be kind enough to point out ? The solution is perfectly clear and correct.
    $endgroup$
    – Rebellos
    Dec 18 '18 at 18:42




    $begingroup$
    The person who disliked would be kind enough to point out ? The solution is perfectly clear and correct.
    $endgroup$
    – Rebellos
    Dec 18 '18 at 18:42











    0












    $begingroup$

    By parts $u=(1+sqrt{x})^3$ and $v'=sqrt{x}$:



    $$int sqrt{x}left(1+sqrt{x}right)^3dx=frac{2}{3}x^{frac{3}{2}}left(1+sqrt{x}right)^3-int :xleft(1+sqrt{x}right)^2dx$$



    And then by expanding:
    $$int :xleft(1+sqrt{x}right)^2dx=frac{x^2}{2}+frac{4}{5}x^{frac{5}{2}}+frac{x^3}{3}$$



    So
    $$int sqrt{x}left(1+sqrt{x}right)^3dx=frac{2}{3}x^{frac{3}{2}}left(1+sqrt{x}right)^3-frac{x^2}{2}-frac{4}{5}x^{frac{5}{2}}-frac{x^3}{3}+C.$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      By parts $u=(1+sqrt{x})^3$ and $v'=sqrt{x}$:



      $$int sqrt{x}left(1+sqrt{x}right)^3dx=frac{2}{3}x^{frac{3}{2}}left(1+sqrt{x}right)^3-int :xleft(1+sqrt{x}right)^2dx$$



      And then by expanding:
      $$int :xleft(1+sqrt{x}right)^2dx=frac{x^2}{2}+frac{4}{5}x^{frac{5}{2}}+frac{x^3}{3}$$



      So
      $$int sqrt{x}left(1+sqrt{x}right)^3dx=frac{2}{3}x^{frac{3}{2}}left(1+sqrt{x}right)^3-frac{x^2}{2}-frac{4}{5}x^{frac{5}{2}}-frac{x^3}{3}+C.$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        By parts $u=(1+sqrt{x})^3$ and $v'=sqrt{x}$:



        $$int sqrt{x}left(1+sqrt{x}right)^3dx=frac{2}{3}x^{frac{3}{2}}left(1+sqrt{x}right)^3-int :xleft(1+sqrt{x}right)^2dx$$



        And then by expanding:
        $$int :xleft(1+sqrt{x}right)^2dx=frac{x^2}{2}+frac{4}{5}x^{frac{5}{2}}+frac{x^3}{3}$$



        So
        $$int sqrt{x}left(1+sqrt{x}right)^3dx=frac{2}{3}x^{frac{3}{2}}left(1+sqrt{x}right)^3-frac{x^2}{2}-frac{4}{5}x^{frac{5}{2}}-frac{x^3}{3}+C.$$






        share|cite|improve this answer









        $endgroup$



        By parts $u=(1+sqrt{x})^3$ and $v'=sqrt{x}$:



        $$int sqrt{x}left(1+sqrt{x}right)^3dx=frac{2}{3}x^{frac{3}{2}}left(1+sqrt{x}right)^3-int :xleft(1+sqrt{x}right)^2dx$$



        And then by expanding:
        $$int :xleft(1+sqrt{x}right)^2dx=frac{x^2}{2}+frac{4}{5}x^{frac{5}{2}}+frac{x^3}{3}$$



        So
        $$int sqrt{x}left(1+sqrt{x}right)^3dx=frac{2}{3}x^{frac{3}{2}}left(1+sqrt{x}right)^3-frac{x^2}{2}-frac{4}{5}x^{frac{5}{2}}-frac{x^3}{3}+C.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 17 '18 at 15:47









        orangeorange

        687315




        687315






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044088%2fsolving-int-sqrtx-left1-sqrtx-right3-mathrmdx%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
            Read more

            Verónica Boquete

            Image
            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
            Read more