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

15.4k31250




15.4k31250










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









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    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

    15.4k31250




    15.4k31250












    • $begingroup$
      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

        689315




        689315






























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