How to evaluate $intfrac{1}{3+4x+4x^2}$?












2












$begingroup$



How to evaluate
$$intfrac{1}{3+4x+4x^2}quad ?
$$




This is what I've done so far:
$$
frac{1}{4} intfrac{1}{x^2+x+frac{3}{4}}
=frac{1}{4} intfrac{1}{(x+frac{1}{2})^2 + frac{1}{2}}
$$



$$
y = frac{1}{a}arctanfrac{u}{a},quad frac{dy}{du} = frac{1}{a^2 + u^2}
$$
$$
a = sqrt{frac{1}{2}},quad
u = (x+frac{1}{2}),quad frac{du}{dx} = 1
$$
so
$$ frac{dy}{dx} = frac{dy}{du} cdot 1
$$



I don't know how to proceed.
Could I also have some help with
$$
intfrac{1}{sqrt{-4x^2-4x+3}}quad ?
$$






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    $$ x^2 + x + frac{3}{4} = left( x + frac{1}{2} right)^2 + frac{1}{2} $$
    $endgroup$
    – Will Jagy
    Dec 10 '18 at 1:07










  • $begingroup$
    Unrelated note: remember to add a $dx$ at the end of your integrals.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:10










  • $begingroup$
    Fixed. Thanks for pointing it out
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:10










  • $begingroup$
    so is the answer $$frac{sqrt2}{4}arctansqrt2({x+frac{1}{2}})$$?
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:20








  • 1




    $begingroup$
    If you want to check the answer, try going to Wolfram Alpha, which can basically calculate everything for you.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:21
















2












$begingroup$



How to evaluate
$$intfrac{1}{3+4x+4x^2}quad ?
$$




This is what I've done so far:
$$
frac{1}{4} intfrac{1}{x^2+x+frac{3}{4}}
=frac{1}{4} intfrac{1}{(x+frac{1}{2})^2 + frac{1}{2}}
$$



$$
y = frac{1}{a}arctanfrac{u}{a},quad frac{dy}{du} = frac{1}{a^2 + u^2}
$$
$$
a = sqrt{frac{1}{2}},quad
u = (x+frac{1}{2}),quad frac{du}{dx} = 1
$$
so
$$ frac{dy}{dx} = frac{dy}{du} cdot 1
$$



I don't know how to proceed.
Could I also have some help with
$$
intfrac{1}{sqrt{-4x^2-4x+3}}quad ?
$$






calculus integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    $$ x^2 + x + frac{3}{4} = left( x + frac{1}{2} right)^2 + frac{1}{2} $$
    $endgroup$
    – Will Jagy
    Dec 10 '18 at 1:07










  • $begingroup$
    Unrelated note: remember to add a $dx$ at the end of your integrals.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:10










  • $begingroup$
    Fixed. Thanks for pointing it out
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:10










  • $begingroup$
    so is the answer $$frac{sqrt2}{4}arctansqrt2({x+frac{1}{2}})$$?
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:20








  • 1




    $begingroup$
    If you want to check the answer, try going to Wolfram Alpha, which can basically calculate everything for you.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:21














2












2








2





$begingroup$



How to evaluate
$$intfrac{1}{3+4x+4x^2}quad ?
$$




This is what I've done so far:
$$
frac{1}{4} intfrac{1}{x^2+x+frac{3}{4}}
=frac{1}{4} intfrac{1}{(x+frac{1}{2})^2 + frac{1}{2}}
$$



$$
y = frac{1}{a}arctanfrac{u}{a},quad frac{dy}{du} = frac{1}{a^2 + u^2}
$$
$$
a = sqrt{frac{1}{2}},quad
u = (x+frac{1}{2}),quad frac{du}{dx} = 1
$$
so
$$ frac{dy}{dx} = frac{dy}{du} cdot 1
$$



I don't know how to proceed.
Could I also have some help with
$$
intfrac{1}{sqrt{-4x^2-4x+3}}quad ?
$$






calculus integration







share|cite|improve this question











$endgroup$





How to evaluate
$$intfrac{1}{3+4x+4x^2}quad ?
$$




This is what I've done so far:
$$
frac{1}{4} intfrac{1}{x^2+x+frac{3}{4}}
=frac{1}{4} intfrac{1}{(x+frac{1}{2})^2 + frac{1}{2}}
$$



$$
y = frac{1}{a}arctanfrac{u}{a},quad frac{dy}{du} = frac{1}{a^2 + u^2}
$$
$$
a = sqrt{frac{1}{2}},quad
u = (x+frac{1}{2}),quad frac{du}{dx} = 1
$$
so
$$ frac{dy}{dx} = frac{dy}{du} cdot 1
$$



I don't know how to proceed.
Could I also have some help with
$$
intfrac{1}{sqrt{-4x^2-4x+3}}quad ?
$$






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 10 '18 at 3:35









user587192

1




1










asked Dec 10 '18 at 1:02









YeeetYeeet

133




133












  • $begingroup$
    $$ x^2 + x + frac{3}{4} = left( x + frac{1}{2} right)^2 + frac{1}{2} $$
    $endgroup$
    – Will Jagy
    Dec 10 '18 at 1:07










  • $begingroup$
    Unrelated note: remember to add a $dx$ at the end of your integrals.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:10










  • $begingroup$
    Fixed. Thanks for pointing it out
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:10










  • $begingroup$
    so is the answer $$frac{sqrt2}{4}arctansqrt2({x+frac{1}{2}})$$?
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:20








  • 1




    $begingroup$
    If you want to check the answer, try going to Wolfram Alpha, which can basically calculate everything for you.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:21


















  • $begingroup$
    $$ x^2 + x + frac{3}{4} = left( x + frac{1}{2} right)^2 + frac{1}{2} $$
    $endgroup$
    – Will Jagy
    Dec 10 '18 at 1:07










  • $begingroup$
    Unrelated note: remember to add a $dx$ at the end of your integrals.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:10










  • $begingroup$
    Fixed. Thanks for pointing it out
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:10










  • $begingroup$
    so is the answer $$frac{sqrt2}{4}arctansqrt2({x+frac{1}{2}})$$?
    $endgroup$
    – Yeeet
    Dec 10 '18 at 1:20








  • 1




    $begingroup$
    If you want to check the answer, try going to Wolfram Alpha, which can basically calculate everything for you.
    $endgroup$
    – YiFan
    Dec 10 '18 at 1:21
















$begingroup$
$$ x^2 + x + frac{3}{4} = left( x + frac{1}{2} right)^2 + frac{1}{2} $$
$endgroup$
– Will Jagy
Dec 10 '18 at 1:07




$begingroup$
$$ x^2 + x + frac{3}{4} = left( x + frac{1}{2} right)^2 + frac{1}{2} $$
$endgroup$
– Will Jagy
Dec 10 '18 at 1:07












$begingroup$
Unrelated note: remember to add a $dx$ at the end of your integrals.
$endgroup$
– YiFan
Dec 10 '18 at 1:10




$begingroup$
Unrelated note: remember to add a $dx$ at the end of your integrals.
$endgroup$
– YiFan
Dec 10 '18 at 1:10












$begingroup$
Fixed. Thanks for pointing it out
$endgroup$
– Yeeet
Dec 10 '18 at 1:10




$begingroup$
Fixed. Thanks for pointing it out
$endgroup$
– Yeeet
Dec 10 '18 at 1:10












$begingroup$
so is the answer $$frac{sqrt2}{4}arctansqrt2({x+frac{1}{2}})$$?
$endgroup$
– Yeeet
Dec 10 '18 at 1:20






$begingroup$
so is the answer $$frac{sqrt2}{4}arctansqrt2({x+frac{1}{2}})$$?
$endgroup$
– Yeeet
Dec 10 '18 at 1:20






1




1




$begingroup$
If you want to check the answer, try going to Wolfram Alpha, which can basically calculate everything for you.
$endgroup$
– YiFan
Dec 10 '18 at 1:21




$begingroup$
If you want to check the answer, try going to Wolfram Alpha, which can basically calculate everything for you.
$endgroup$
– YiFan
Dec 10 '18 at 1:21










2 Answers
2






active

oldest

votes


















3












$begingroup$

What can be done is a double U-Substitution



We have $$intfrac{1}{3+4x+4x^2}dx$$



By completing the square we see $$intfrac{1}{3+4x+4x^2}dx = intfrac{1}{(2x+1)^2+2}dx$$



Now substitute $u = 2x+1$ and $du = 2udx$. Now our integral becomes $$frac{1}{2}intfrac{1}{u^2+2}du = frac{1}{2}intfrac{1}{2(frac{u^2}{2}+1)}du = frac{1}{4}intfrac{1}{(frac{u^2}{2}+1)}du$$



Now we substitute $s = frac{u}{sqrt{2}}$ and $ds = frac{1}{sqrt{2}}du$
in the integrand. $$frac{1}{2sqrt{2}}intfrac{1}{s^2+1}ds$$



$$intfrac{1}{s^2+1}ds = tan^{-1}(s)$$ So we are left with $$frac{tan^{-1}(s)}{2sqrt{2}}$$



Back substituting $s = frac{u}{sqrt{2}}$, $$frac{tan^{-1}(s)}{2sqrt{2}} = frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}}$$



Now back substitute $u = 2x+1$ and $$frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}} = frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}$$,



so our final answer is $$boxed{frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}+ C}$$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    HINT:Complete the squares and use the results



    $$intfrac{1}{x^2 + a^2}dx= frac{1}{a}arctan frac{x}{a}+C $$
    and
    $$intfrac{1}{sqrt{a^2-x^2}}dx=arcsin frac{x}{a}+C$$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      What can be done is a double U-Substitution



      We have $$intfrac{1}{3+4x+4x^2}dx$$



      By completing the square we see $$intfrac{1}{3+4x+4x^2}dx = intfrac{1}{(2x+1)^2+2}dx$$



      Now substitute $u = 2x+1$ and $du = 2udx$. Now our integral becomes $$frac{1}{2}intfrac{1}{u^2+2}du = frac{1}{2}intfrac{1}{2(frac{u^2}{2}+1)}du = frac{1}{4}intfrac{1}{(frac{u^2}{2}+1)}du$$



      Now we substitute $s = frac{u}{sqrt{2}}$ and $ds = frac{1}{sqrt{2}}du$
      in the integrand. $$frac{1}{2sqrt{2}}intfrac{1}{s^2+1}ds$$



      $$intfrac{1}{s^2+1}ds = tan^{-1}(s)$$ So we are left with $$frac{tan^{-1}(s)}{2sqrt{2}}$$



      Back substituting $s = frac{u}{sqrt{2}}$, $$frac{tan^{-1}(s)}{2sqrt{2}} = frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}}$$



      Now back substitute $u = 2x+1$ and $$frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}} = frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}$$,



      so our final answer is $$boxed{frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}+ C}$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        What can be done is a double U-Substitution



        We have $$intfrac{1}{3+4x+4x^2}dx$$



        By completing the square we see $$intfrac{1}{3+4x+4x^2}dx = intfrac{1}{(2x+1)^2+2}dx$$



        Now substitute $u = 2x+1$ and $du = 2udx$. Now our integral becomes $$frac{1}{2}intfrac{1}{u^2+2}du = frac{1}{2}intfrac{1}{2(frac{u^2}{2}+1)}du = frac{1}{4}intfrac{1}{(frac{u^2}{2}+1)}du$$



        Now we substitute $s = frac{u}{sqrt{2}}$ and $ds = frac{1}{sqrt{2}}du$
        in the integrand. $$frac{1}{2sqrt{2}}intfrac{1}{s^2+1}ds$$



        $$intfrac{1}{s^2+1}ds = tan^{-1}(s)$$ So we are left with $$frac{tan^{-1}(s)}{2sqrt{2}}$$



        Back substituting $s = frac{u}{sqrt{2}}$, $$frac{tan^{-1}(s)}{2sqrt{2}} = frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}}$$



        Now back substitute $u = 2x+1$ and $$frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}} = frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}$$,



        so our final answer is $$boxed{frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}+ C}$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          What can be done is a double U-Substitution



          We have $$intfrac{1}{3+4x+4x^2}dx$$



          By completing the square we see $$intfrac{1}{3+4x+4x^2}dx = intfrac{1}{(2x+1)^2+2}dx$$



          Now substitute $u = 2x+1$ and $du = 2udx$. Now our integral becomes $$frac{1}{2}intfrac{1}{u^2+2}du = frac{1}{2}intfrac{1}{2(frac{u^2}{2}+1)}du = frac{1}{4}intfrac{1}{(frac{u^2}{2}+1)}du$$



          Now we substitute $s = frac{u}{sqrt{2}}$ and $ds = frac{1}{sqrt{2}}du$
          in the integrand. $$frac{1}{2sqrt{2}}intfrac{1}{s^2+1}ds$$



          $$intfrac{1}{s^2+1}ds = tan^{-1}(s)$$ So we are left with $$frac{tan^{-1}(s)}{2sqrt{2}}$$



          Back substituting $s = frac{u}{sqrt{2}}$, $$frac{tan^{-1}(s)}{2sqrt{2}} = frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}}$$



          Now back substitute $u = 2x+1$ and $$frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}} = frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}$$,



          so our final answer is $$boxed{frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}+ C}$$






          share|cite|improve this answer









          $endgroup$



          What can be done is a double U-Substitution



          We have $$intfrac{1}{3+4x+4x^2}dx$$



          By completing the square we see $$intfrac{1}{3+4x+4x^2}dx = intfrac{1}{(2x+1)^2+2}dx$$



          Now substitute $u = 2x+1$ and $du = 2udx$. Now our integral becomes $$frac{1}{2}intfrac{1}{u^2+2}du = frac{1}{2}intfrac{1}{2(frac{u^2}{2}+1)}du = frac{1}{4}intfrac{1}{(frac{u^2}{2}+1)}du$$



          Now we substitute $s = frac{u}{sqrt{2}}$ and $ds = frac{1}{sqrt{2}}du$
          in the integrand. $$frac{1}{2sqrt{2}}intfrac{1}{s^2+1}ds$$



          $$intfrac{1}{s^2+1}ds = tan^{-1}(s)$$ So we are left with $$frac{tan^{-1}(s)}{2sqrt{2}}$$



          Back substituting $s = frac{u}{sqrt{2}}$, $$frac{tan^{-1}(s)}{2sqrt{2}} = frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}}$$



          Now back substitute $u = 2x+1$ and $$frac{tan^{-1}(frac{u}{sqrt{2}})}{2sqrt{2}} = frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}$$,



          so our final answer is $$boxed{frac{tan^{-1}(frac{2x+1}{sqrt{2}})}{2sqrt{2}}+ C}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 1:29









          Aniruddh VenkatesanAniruddh Venkatesan

          149112




          149112























              3












              $begingroup$

              HINT:Complete the squares and use the results



              $$intfrac{1}{x^2 + a^2}dx= frac{1}{a}arctan frac{x}{a}+C $$
              and
              $$intfrac{1}{sqrt{a^2-x^2}}dx=arcsin frac{x}{a}+C$$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                HINT:Complete the squares and use the results



                $$intfrac{1}{x^2 + a^2}dx= frac{1}{a}arctan frac{x}{a}+C $$
                and
                $$intfrac{1}{sqrt{a^2-x^2}}dx=arcsin frac{x}{a}+C$$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  HINT:Complete the squares and use the results



                  $$intfrac{1}{x^2 + a^2}dx= frac{1}{a}arctan frac{x}{a}+C $$
                  and
                  $$intfrac{1}{sqrt{a^2-x^2}}dx=arcsin frac{x}{a}+C$$.






                  share|cite|improve this answer









                  $endgroup$



                  HINT:Complete the squares and use the results



                  $$intfrac{1}{x^2 + a^2}dx= frac{1}{a}arctan frac{x}{a}+C $$
                  and
                  $$intfrac{1}{sqrt{a^2-x^2}}dx=arcsin frac{x}{a}+C$$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 1:31









                  Thomas ShelbyThomas Shelby

                  3,3271524




                  3,3271524






























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                      Ida-Boy-Ed-Garten

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                      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
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