Is integration by parts the best method?












0












$begingroup$


My integration techniques are rusty. If I wish to integrate the following function, I can do it by using integration by parts twice, but is this really the simplest way?



I wish to integrate the function $f(x) = (x^2 - 2x + 1)(e^{-x})$. Integrating by parts we get $$int f(x)dx = (x^2 - 2x + 1)(-e^{-x}) - int(2x-2)(-e^{-x})dx$$
In order to solve the last integral we must again do integration by parts to get $$int(2x-2)(-e^{-x})dx = (2x-2)(e^{-x})-int 2(e^{-x})dx$$
The final integral can then be solved directly and we get the final result, which is $$int f(x)dx = -e^{-x}(x^2+1)$$
But is this the simplest way? Does this mean I need to use integration by parts $n$ times if I have a polynomium of $n$th degree? Or are there some tricks when a polynomium is multiplied by $e^{-x}$?






integration







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  • 1




    $begingroup$
    The only "trick" I can think of is using a reduction formula but that's derived from integration by parts.
    $endgroup$
    – Justin Stevenson
    Dec 10 '18 at 1:28
















0












$begingroup$


My integration techniques are rusty. If I wish to integrate the following function, I can do it by using integration by parts twice, but is this really the simplest way?



I wish to integrate the function $f(x) = (x^2 - 2x + 1)(e^{-x})$. Integrating by parts we get $$int f(x)dx = (x^2 - 2x + 1)(-e^{-x}) - int(2x-2)(-e^{-x})dx$$
In order to solve the last integral we must again do integration by parts to get $$int(2x-2)(-e^{-x})dx = (2x-2)(e^{-x})-int 2(e^{-x})dx$$
The final integral can then be solved directly and we get the final result, which is $$int f(x)dx = -e^{-x}(x^2+1)$$
But is this the simplest way? Does this mean I need to use integration by parts $n$ times if I have a polynomium of $n$th degree? Or are there some tricks when a polynomium is multiplied by $e^{-x}$?






integration







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The only "trick" I can think of is using a reduction formula but that's derived from integration by parts.
    $endgroup$
    – Justin Stevenson
    Dec 10 '18 at 1:28














0












0








0





$begingroup$


My integration techniques are rusty. If I wish to integrate the following function, I can do it by using integration by parts twice, but is this really the simplest way?



I wish to integrate the function $f(x) = (x^2 - 2x + 1)(e^{-x})$. Integrating by parts we get $$int f(x)dx = (x^2 - 2x + 1)(-e^{-x}) - int(2x-2)(-e^{-x})dx$$
In order to solve the last integral we must again do integration by parts to get $$int(2x-2)(-e^{-x})dx = (2x-2)(e^{-x})-int 2(e^{-x})dx$$
The final integral can then be solved directly and we get the final result, which is $$int f(x)dx = -e^{-x}(x^2+1)$$
But is this the simplest way? Does this mean I need to use integration by parts $n$ times if I have a polynomium of $n$th degree? Or are there some tricks when a polynomium is multiplied by $e^{-x}$?






integration







share|cite|improve this question









$endgroup$




My integration techniques are rusty. If I wish to integrate the following function, I can do it by using integration by parts twice, but is this really the simplest way?



I wish to integrate the function $f(x) = (x^2 - 2x + 1)(e^{-x})$. Integrating by parts we get $$int f(x)dx = (x^2 - 2x + 1)(-e^{-x}) - int(2x-2)(-e^{-x})dx$$
In order to solve the last integral we must again do integration by parts to get $$int(2x-2)(-e^{-x})dx = (2x-2)(e^{-x})-int 2(e^{-x})dx$$
The final integral can then be solved directly and we get the final result, which is $$int f(x)dx = -e^{-x}(x^2+1)$$
But is this the simplest way? Does this mean I need to use integration by parts $n$ times if I have a polynomium of $n$th degree? Or are there some tricks when a polynomium is multiplied by $e^{-x}$?






integration




integration






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asked Dec 10 '18 at 1:01









JensJens

3,79021030




3,79021030








  • 1




    $begingroup$
    The only "trick" I can think of is using a reduction formula but that's derived from integration by parts.
    $endgroup$
    – Justin Stevenson
    Dec 10 '18 at 1:28














  • 1




    $begingroup$
    The only "trick" I can think of is using a reduction formula but that's derived from integration by parts.
    $endgroup$
    – Justin Stevenson
    Dec 10 '18 at 1:28








1




1




$begingroup$
The only "trick" I can think of is using a reduction formula but that's derived from integration by parts.
$endgroup$
– Justin Stevenson
Dec 10 '18 at 1:28




$begingroup$
The only "trick" I can think of is using a reduction formula but that's derived from integration by parts.
$endgroup$
– Justin Stevenson
Dec 10 '18 at 1:28










3 Answers
3






active

oldest

votes


















0












$begingroup$

A trick for such integrals:




  • If $f(x)$ is of the form $f(x)$ = $e^x$ $(g(x) + gprime(x))$, then
    $int f(x)$ = $e^x g(x)$ + some constant.

  • If $f(x)$ is of the form $f(x)$ = $e^{-x}$ $(gprime(x) - g(x))$,
    then $int f(x)$ = $e^{-x}g(x)$ + some constant.


The given question is of the second form.
I am sure you can prove the trick so I am skipping it.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Integration by parts done fast
    This method speeds up the process, but other then this I don't know of any 'tricks'






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      "Best method" is subjective. But another way is to use Feynman's trick of differentiating under the integral sign.



      First, note that the integral is $displaystyle frac 1e int (x-1)^2e^{-(x-1)}dx$



      Now consider the integral $g(x,k) = displaystyle frac 1e int e^{-k(x-1)}dx$ and note that $displaystyle frac{partial{^2}}{partial k^2}g(x,k) = frac 1e int (x-1)^2e^{-(x-1)}dx$ when $k=1$.



      $g(x,k)$ is trivial to compute, and the twice partial differentiation can be done with product rule without too much fuss.



      EDIT (addendum): even if you were to use integration by parts twice, your life would be made easier by doing a substitution $displaystyle u = x-1$, to make the integral $displaystyle frac 1e int u^2e^{-u} du$, which is more tractable.






      share|cite|improve this answer











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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        A trick for such integrals:




        • If $f(x)$ is of the form $f(x)$ = $e^x$ $(g(x) + gprime(x))$, then
          $int f(x)$ = $e^x g(x)$ + some constant.

        • If $f(x)$ is of the form $f(x)$ = $e^{-x}$ $(gprime(x) - g(x))$,
          then $int f(x)$ = $e^{-x}g(x)$ + some constant.


        The given question is of the second form.
        I am sure you can prove the trick so I am skipping it.






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          A trick for such integrals:




          • If $f(x)$ is of the form $f(x)$ = $e^x$ $(g(x) + gprime(x))$, then
            $int f(x)$ = $e^x g(x)$ + some constant.

          • If $f(x)$ is of the form $f(x)$ = $e^{-x}$ $(gprime(x) - g(x))$,
            then $int f(x)$ = $e^{-x}g(x)$ + some constant.


          The given question is of the second form.
          I am sure you can prove the trick so I am skipping it.






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            A trick for such integrals:




            • If $f(x)$ is of the form $f(x)$ = $e^x$ $(g(x) + gprime(x))$, then
              $int f(x)$ = $e^x g(x)$ + some constant.

            • If $f(x)$ is of the form $f(x)$ = $e^{-x}$ $(gprime(x) - g(x))$,
              then $int f(x)$ = $e^{-x}g(x)$ + some constant.


            The given question is of the second form.
            I am sure you can prove the trick so I am skipping it.






            share|cite|improve this answer









            $endgroup$



            A trick for such integrals:




            • If $f(x)$ is of the form $f(x)$ = $e^x$ $(g(x) + gprime(x))$, then
              $int f(x)$ = $e^x g(x)$ + some constant.

            • If $f(x)$ is of the form $f(x)$ = $e^{-x}$ $(gprime(x) - g(x))$,
              then $int f(x)$ = $e^{-x}g(x)$ + some constant.


            The given question is of the second form.
            I am sure you can prove the trick so I am skipping it.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 10 '18 at 2:48









            Nishant Pr. DasNishant Pr. Das

            234




            234























                1












                $begingroup$

                Integration by parts done fast
                This method speeds up the process, but other then this I don't know of any 'tricks'






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Integration by parts done fast
                  This method speeds up the process, but other then this I don't know of any 'tricks'






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Integration by parts done fast
                    This method speeds up the process, but other then this I don't know of any 'tricks'






                    share|cite|improve this answer









                    $endgroup$



                    Integration by parts done fast
                    This method speeds up the process, but other then this I don't know of any 'tricks'







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 10 '18 at 1:40









                    rickyricky

                    1207




                    1207























                        1












                        $begingroup$

                        "Best method" is subjective. But another way is to use Feynman's trick of differentiating under the integral sign.



                        First, note that the integral is $displaystyle frac 1e int (x-1)^2e^{-(x-1)}dx$



                        Now consider the integral $g(x,k) = displaystyle frac 1e int e^{-k(x-1)}dx$ and note that $displaystyle frac{partial{^2}}{partial k^2}g(x,k) = frac 1e int (x-1)^2e^{-(x-1)}dx$ when $k=1$.



                        $g(x,k)$ is trivial to compute, and the twice partial differentiation can be done with product rule without too much fuss.



                        EDIT (addendum): even if you were to use integration by parts twice, your life would be made easier by doing a substitution $displaystyle u = x-1$, to make the integral $displaystyle frac 1e int u^2e^{-u} du$, which is more tractable.






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          "Best method" is subjective. But another way is to use Feynman's trick of differentiating under the integral sign.



                          First, note that the integral is $displaystyle frac 1e int (x-1)^2e^{-(x-1)}dx$



                          Now consider the integral $g(x,k) = displaystyle frac 1e int e^{-k(x-1)}dx$ and note that $displaystyle frac{partial{^2}}{partial k^2}g(x,k) = frac 1e int (x-1)^2e^{-(x-1)}dx$ when $k=1$.



                          $g(x,k)$ is trivial to compute, and the twice partial differentiation can be done with product rule without too much fuss.



                          EDIT (addendum): even if you were to use integration by parts twice, your life would be made easier by doing a substitution $displaystyle u = x-1$, to make the integral $displaystyle frac 1e int u^2e^{-u} du$, which is more tractable.






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            "Best method" is subjective. But another way is to use Feynman's trick of differentiating under the integral sign.



                            First, note that the integral is $displaystyle frac 1e int (x-1)^2e^{-(x-1)}dx$



                            Now consider the integral $g(x,k) = displaystyle frac 1e int e^{-k(x-1)}dx$ and note that $displaystyle frac{partial{^2}}{partial k^2}g(x,k) = frac 1e int (x-1)^2e^{-(x-1)}dx$ when $k=1$.



                            $g(x,k)$ is trivial to compute, and the twice partial differentiation can be done with product rule without too much fuss.



                            EDIT (addendum): even if you were to use integration by parts twice, your life would be made easier by doing a substitution $displaystyle u = x-1$, to make the integral $displaystyle frac 1e int u^2e^{-u} du$, which is more tractable.






                            share|cite|improve this answer











                            $endgroup$



                            "Best method" is subjective. But another way is to use Feynman's trick of differentiating under the integral sign.



                            First, note that the integral is $displaystyle frac 1e int (x-1)^2e^{-(x-1)}dx$



                            Now consider the integral $g(x,k) = displaystyle frac 1e int e^{-k(x-1)}dx$ and note that $displaystyle frac{partial{^2}}{partial k^2}g(x,k) = frac 1e int (x-1)^2e^{-(x-1)}dx$ when $k=1$.



                            $g(x,k)$ is trivial to compute, and the twice partial differentiation can be done with product rule without too much fuss.



                            EDIT (addendum): even if you were to use integration by parts twice, your life would be made easier by doing a substitution $displaystyle u = x-1$, to make the integral $displaystyle frac 1e int u^2e^{-u} du$, which is more tractable.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 10 '18 at 1:52

























                            answered Dec 10 '18 at 1:42









                            DeepakDeepak

                            17k11536




                            17k11536






























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