Integral $int_0^1 frac{sqrt{x}}{3+x}; dx$ [closed]











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How can we compute integration below?



$$int_0^1 frac{sqrt{x}}{3+x} ;dx$$



Solution shows the answer is 2-π/3 but I can not derive this equation.






integration







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closed as off-topic by Did, Gibbs, Leucippus, Scientifica, Anurag A Nov 16 at 3:32


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Gibbs, Leucippus, Scientifica, Anurag A

If this question can be reworded to fit the rules in the help center, please edit the question.













  • HINT: Change of variables $x=t^2$
    – Tito Eliatron
    Nov 15 at 20:34












  • And even, $$x=3t^2$$
    – Did
    Nov 15 at 20:36















up vote
-2
down vote

favorite












How can we compute integration below?



$$int_0^1 frac{sqrt{x}}{3+x} ;dx$$



Solution shows the answer is 2-π/3 but I can not derive this equation.






integration







share|cite|improve this question















closed as off-topic by Did, Gibbs, Leucippus, Scientifica, Anurag A Nov 16 at 3:32


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Gibbs, Leucippus, Scientifica, Anurag A

If this question can be reworded to fit the rules in the help center, please edit the question.













  • HINT: Change of variables $x=t^2$
    – Tito Eliatron
    Nov 15 at 20:34












  • And even, $$x=3t^2$$
    – Did
    Nov 15 at 20:36













up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











How can we compute integration below?



$$int_0^1 frac{sqrt{x}}{3+x} ;dx$$



Solution shows the answer is 2-π/3 but I can not derive this equation.






integration







share|cite|improve this question















How can we compute integration below?



$$int_0^1 frac{sqrt{x}}{3+x} ;dx$$



Solution shows the answer is 2-π/3 but I can not derive this equation.






integration




integration






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edited Nov 15 at 20:34









MPW

29.4k11856




29.4k11856










asked Nov 15 at 20:29









owen

91




91




closed as off-topic by Did, Gibbs, Leucippus, Scientifica, Anurag A Nov 16 at 3:32


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Gibbs, Leucippus, Scientifica, Anurag A

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Did, Gibbs, Leucippus, Scientifica, Anurag A Nov 16 at 3:32


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Gibbs, Leucippus, Scientifica, Anurag A

If this question can be reworded to fit the rules in the help center, please edit the question.












  • HINT: Change of variables $x=t^2$
    – Tito Eliatron
    Nov 15 at 20:34












  • And even, $$x=3t^2$$
    – Did
    Nov 15 at 20:36


















  • HINT: Change of variables $x=t^2$
    – Tito Eliatron
    Nov 15 at 20:34












  • And even, $$x=3t^2$$
    – Did
    Nov 15 at 20:36
















HINT: Change of variables $x=t^2$
– Tito Eliatron
Nov 15 at 20:34






HINT: Change of variables $x=t^2$
– Tito Eliatron
Nov 15 at 20:34














And even, $$x=3t^2$$
– Did
Nov 15 at 20:36




And even, $$x=3t^2$$
– Did
Nov 15 at 20:36










2 Answers
2






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1
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accepted










Make the substitution: $x=u^2, , dx=2u,du$



$$int_0^1 frac{sqrt{x}}{3+x} dx=2int_0^1 frac{u^2}{3+u^2} du={displaystyleint_0^1}left(dfrac{class{steps-node}{cssId{steps-node-4}{u^2+3}}}{u^2+3}-dfrac{class{steps-node}{cssId{steps-node-5}{3}}}{u^2+3}right)mathrm{d}u={displaystyleint_0^1}left(1-dfrac{3}{u^2+3}right)mathrm{d}u$$
$$={displaystyleint_0^1}1,mathrm{d}u-class{steps-node}{cssId{steps-node-6}{3}}{displaystyleint_0^1}dfrac{1}{u^2+3},mathrm{d}u=1-class{steps-node}{cssId{steps-node-7}{dfrac{1}{sqrt{3}}}}{displaystyleint_0^1}dfrac{1}{v^2+1},mathrm{d}v=1-sqrt{3},arctanleft(frac{u}{sqrt{3}}right)Bigg|_0^1$$
$$=1-sqrt{3}arctanleft(frac{1}{sqrt{3}}right)=1-frac{sqrt{3}pi}{6}=frac{6-sqrt{3}pi}{6}$$






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    Hint:



    Set $t^2=x,;tge 0$ to obtain the integral of a rational function:
    $;mathrm d x=2t,mathrm d t$, $;x=0leftrightarrow t=0$, $;x=1leftrightarrow t=1$, so
    $$int_0^1 frac{sqrt{x}}{3+x} ,mathrm d x=int_0^1 frac{2t^2}{t^2+3} ,mathrm d t = 2int_0^1 Bigl(1-frac{3}{t^2+3} Bigr),mathrm d t=dotsm$$






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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      Make the substitution: $x=u^2, , dx=2u,du$



      $$int_0^1 frac{sqrt{x}}{3+x} dx=2int_0^1 frac{u^2}{3+u^2} du={displaystyleint_0^1}left(dfrac{class{steps-node}{cssId{steps-node-4}{u^2+3}}}{u^2+3}-dfrac{class{steps-node}{cssId{steps-node-5}{3}}}{u^2+3}right)mathrm{d}u={displaystyleint_0^1}left(1-dfrac{3}{u^2+3}right)mathrm{d}u$$
      $$={displaystyleint_0^1}1,mathrm{d}u-class{steps-node}{cssId{steps-node-6}{3}}{displaystyleint_0^1}dfrac{1}{u^2+3},mathrm{d}u=1-class{steps-node}{cssId{steps-node-7}{dfrac{1}{sqrt{3}}}}{displaystyleint_0^1}dfrac{1}{v^2+1},mathrm{d}v=1-sqrt{3},arctanleft(frac{u}{sqrt{3}}right)Bigg|_0^1$$
      $$=1-sqrt{3}arctanleft(frac{1}{sqrt{3}}right)=1-frac{sqrt{3}pi}{6}=frac{6-sqrt{3}pi}{6}$$






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted










        Make the substitution: $x=u^2, , dx=2u,du$



        $$int_0^1 frac{sqrt{x}}{3+x} dx=2int_0^1 frac{u^2}{3+u^2} du={displaystyleint_0^1}left(dfrac{class{steps-node}{cssId{steps-node-4}{u^2+3}}}{u^2+3}-dfrac{class{steps-node}{cssId{steps-node-5}{3}}}{u^2+3}right)mathrm{d}u={displaystyleint_0^1}left(1-dfrac{3}{u^2+3}right)mathrm{d}u$$
        $$={displaystyleint_0^1}1,mathrm{d}u-class{steps-node}{cssId{steps-node-6}{3}}{displaystyleint_0^1}dfrac{1}{u^2+3},mathrm{d}u=1-class{steps-node}{cssId{steps-node-7}{dfrac{1}{sqrt{3}}}}{displaystyleint_0^1}dfrac{1}{v^2+1},mathrm{d}v=1-sqrt{3},arctanleft(frac{u}{sqrt{3}}right)Bigg|_0^1$$
        $$=1-sqrt{3}arctanleft(frac{1}{sqrt{3}}right)=1-frac{sqrt{3}pi}{6}=frac{6-sqrt{3}pi}{6}$$






        share|cite|improve this answer























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Make the substitution: $x=u^2, , dx=2u,du$



          $$int_0^1 frac{sqrt{x}}{3+x} dx=2int_0^1 frac{u^2}{3+u^2} du={displaystyleint_0^1}left(dfrac{class{steps-node}{cssId{steps-node-4}{u^2+3}}}{u^2+3}-dfrac{class{steps-node}{cssId{steps-node-5}{3}}}{u^2+3}right)mathrm{d}u={displaystyleint_0^1}left(1-dfrac{3}{u^2+3}right)mathrm{d}u$$
          $$={displaystyleint_0^1}1,mathrm{d}u-class{steps-node}{cssId{steps-node-6}{3}}{displaystyleint_0^1}dfrac{1}{u^2+3},mathrm{d}u=1-class{steps-node}{cssId{steps-node-7}{dfrac{1}{sqrt{3}}}}{displaystyleint_0^1}dfrac{1}{v^2+1},mathrm{d}v=1-sqrt{3},arctanleft(frac{u}{sqrt{3}}right)Bigg|_0^1$$
          $$=1-sqrt{3}arctanleft(frac{1}{sqrt{3}}right)=1-frac{sqrt{3}pi}{6}=frac{6-sqrt{3}pi}{6}$$






          share|cite|improve this answer












          Make the substitution: $x=u^2, , dx=2u,du$



          $$int_0^1 frac{sqrt{x}}{3+x} dx=2int_0^1 frac{u^2}{3+u^2} du={displaystyleint_0^1}left(dfrac{class{steps-node}{cssId{steps-node-4}{u^2+3}}}{u^2+3}-dfrac{class{steps-node}{cssId{steps-node-5}{3}}}{u^2+3}right)mathrm{d}u={displaystyleint_0^1}left(1-dfrac{3}{u^2+3}right)mathrm{d}u$$
          $$={displaystyleint_0^1}1,mathrm{d}u-class{steps-node}{cssId{steps-node-6}{3}}{displaystyleint_0^1}dfrac{1}{u^2+3},mathrm{d}u=1-class{steps-node}{cssId{steps-node-7}{dfrac{1}{sqrt{3}}}}{displaystyleint_0^1}dfrac{1}{v^2+1},mathrm{d}v=1-sqrt{3},arctanleft(frac{u}{sqrt{3}}right)Bigg|_0^1$$
          $$=1-sqrt{3}arctanleft(frac{1}{sqrt{3}}right)=1-frac{sqrt{3}pi}{6}=frac{6-sqrt{3}pi}{6}$$







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          answered Nov 15 at 21:30









          Jevaut

          5049




          5049






















              up vote
              1
              down vote













              Hint:



              Set $t^2=x,;tge 0$ to obtain the integral of a rational function:
              $;mathrm d x=2t,mathrm d t$, $;x=0leftrightarrow t=0$, $;x=1leftrightarrow t=1$, so
              $$int_0^1 frac{sqrt{x}}{3+x} ,mathrm d x=int_0^1 frac{2t^2}{t^2+3} ,mathrm d t = 2int_0^1 Bigl(1-frac{3}{t^2+3} Bigr),mathrm d t=dotsm$$






              share|cite|improve this answer

























                up vote
                1
                down vote













                Hint:



                Set $t^2=x,;tge 0$ to obtain the integral of a rational function:
                $;mathrm d x=2t,mathrm d t$, $;x=0leftrightarrow t=0$, $;x=1leftrightarrow t=1$, so
                $$int_0^1 frac{sqrt{x}}{3+x} ,mathrm d x=int_0^1 frac{2t^2}{t^2+3} ,mathrm d t = 2int_0^1 Bigl(1-frac{3}{t^2+3} Bigr),mathrm d t=dotsm$$






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  Hint:



                  Set $t^2=x,;tge 0$ to obtain the integral of a rational function:
                  $;mathrm d x=2t,mathrm d t$, $;x=0leftrightarrow t=0$, $;x=1leftrightarrow t=1$, so
                  $$int_0^1 frac{sqrt{x}}{3+x} ,mathrm d x=int_0^1 frac{2t^2}{t^2+3} ,mathrm d t = 2int_0^1 Bigl(1-frac{3}{t^2+3} Bigr),mathrm d t=dotsm$$






                  share|cite|improve this answer












                  Hint:



                  Set $t^2=x,;tge 0$ to obtain the integral of a rational function:
                  $;mathrm d x=2t,mathrm d t$, $;x=0leftrightarrow t=0$, $;x=1leftrightarrow t=1$, so
                  $$int_0^1 frac{sqrt{x}}{3+x} ,mathrm d x=int_0^1 frac{2t^2}{t^2+3} ,mathrm d t = 2int_0^1 Bigl(1-frac{3}{t^2+3} Bigr),mathrm d t=dotsm$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 15 at 21:13









                  Bernard

                  115k637108




                  115k637108















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