Finding the definite integral $int_1^e frac{dx}{xsqrt{1+ln^2x}}$











up vote
6
down vote

favorite
3












So I have the following problem:



$$int_1^{e} frac{1}{xsqrt{1+ln^2x}}dx $$



Can somebody comfirm that the integral of this is



$$ln|sqrt{1+ln^2x}+ ln x|+C$$



and I that the anwser is $$ln |sqrt{2}+1|$$
that is aproximately 0.8814



Does anyone else got the same anwser?










share|cite|improve this question




















  • 3




    Note that the derivative and the integral are inverse operations of each other... at least up to a constant. If you have the antiderivative, you simply can take the derivative of the antiderivative to check if it is correct.
    – Decaf-Math
    Nov 18 at 0:32






  • 1




    You did an indefinite integral. You would have to plug in 1 and e as bounds to get what you actually want.
    – AHusain
    Nov 18 at 0:34










  • Differentiate your ans, with respect to x.
    – John Nash
    Nov 18 at 0:39










  • I edited my answer to include the definite integral. click the check mark next to the voting button if my answer is helpful
    – clathratus
    2 days ago















up vote
6
down vote

favorite
3












So I have the following problem:



$$int_1^{e} frac{1}{xsqrt{1+ln^2x}}dx $$



Can somebody comfirm that the integral of this is



$$ln|sqrt{1+ln^2x}+ ln x|+C$$



and I that the anwser is $$ln |sqrt{2}+1|$$
that is aproximately 0.8814



Does anyone else got the same anwser?










share|cite|improve this question




















  • 3




    Note that the derivative and the integral are inverse operations of each other... at least up to a constant. If you have the antiderivative, you simply can take the derivative of the antiderivative to check if it is correct.
    – Decaf-Math
    Nov 18 at 0:32






  • 1




    You did an indefinite integral. You would have to plug in 1 and e as bounds to get what you actually want.
    – AHusain
    Nov 18 at 0:34










  • Differentiate your ans, with respect to x.
    – John Nash
    Nov 18 at 0:39










  • I edited my answer to include the definite integral. click the check mark next to the voting button if my answer is helpful
    – clathratus
    2 days ago













up vote
6
down vote

favorite
3









up vote
6
down vote

favorite
3






3





So I have the following problem:



$$int_1^{e} frac{1}{xsqrt{1+ln^2x}}dx $$



Can somebody comfirm that the integral of this is



$$ln|sqrt{1+ln^2x}+ ln x|+C$$



and I that the anwser is $$ln |sqrt{2}+1|$$
that is aproximately 0.8814



Does anyone else got the same anwser?










share|cite|improve this question















So I have the following problem:



$$int_1^{e} frac{1}{xsqrt{1+ln^2x}}dx $$



Can somebody comfirm that the integral of this is



$$ln|sqrt{1+ln^2x}+ ln x|+C$$



and I that the anwser is $$ln |sqrt{2}+1|$$
that is aproximately 0.8814



Does anyone else got the same anwser?







calculus integration definite-integrals closed-form






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 at 10:22

























asked Nov 18 at 0:30









Student123

363




363








  • 3




    Note that the derivative and the integral are inverse operations of each other... at least up to a constant. If you have the antiderivative, you simply can take the derivative of the antiderivative to check if it is correct.
    – Decaf-Math
    Nov 18 at 0:32






  • 1




    You did an indefinite integral. You would have to plug in 1 and e as bounds to get what you actually want.
    – AHusain
    Nov 18 at 0:34










  • Differentiate your ans, with respect to x.
    – John Nash
    Nov 18 at 0:39










  • I edited my answer to include the definite integral. click the check mark next to the voting button if my answer is helpful
    – clathratus
    2 days ago














  • 3




    Note that the derivative and the integral are inverse operations of each other... at least up to a constant. If you have the antiderivative, you simply can take the derivative of the antiderivative to check if it is correct.
    – Decaf-Math
    Nov 18 at 0:32






  • 1




    You did an indefinite integral. You would have to plug in 1 and e as bounds to get what you actually want.
    – AHusain
    Nov 18 at 0:34










  • Differentiate your ans, with respect to x.
    – John Nash
    Nov 18 at 0:39










  • I edited my answer to include the definite integral. click the check mark next to the voting button if my answer is helpful
    – clathratus
    2 days ago








3




3




Note that the derivative and the integral are inverse operations of each other... at least up to a constant. If you have the antiderivative, you simply can take the derivative of the antiderivative to check if it is correct.
– Decaf-Math
Nov 18 at 0:32




Note that the derivative and the integral are inverse operations of each other... at least up to a constant. If you have the antiderivative, you simply can take the derivative of the antiderivative to check if it is correct.
– Decaf-Math
Nov 18 at 0:32




1




1




You did an indefinite integral. You would have to plug in 1 and e as bounds to get what you actually want.
– AHusain
Nov 18 at 0:34




You did an indefinite integral. You would have to plug in 1 and e as bounds to get what you actually want.
– AHusain
Nov 18 at 0:34












Differentiate your ans, with respect to x.
– John Nash
Nov 18 at 0:39




Differentiate your ans, with respect to x.
– John Nash
Nov 18 at 0:39












I edited my answer to include the definite integral. click the check mark next to the voting button if my answer is helpful
– clathratus
2 days ago




I edited my answer to include the definite integral. click the check mark next to the voting button if my answer is helpful
– clathratus
2 days ago










2 Answers
2






active

oldest

votes

















up vote
7
down vote













Hint. One may perform the change of variable
$$
u=ln x,qquad du=frac{dx}x,
$$
giving
$$
int frac{1}{xsqrt{1+ln^2x}}:dx=int frac{du}{sqrt{1+u^2}}
$$
then one may notice that
$$
left[ln left(u+ sqrt{1+u^2}right)right]'=frac{1+frac{u}{sqrt{1+u^2}}}{u+sqrt{1+u^2}}=frac{1}{sqrt{1+u^2}}.
$$






share|cite|improve this answer





















  • thank you so much for helping! but i have a question: before you change the variable there is a 'x' in front of the square root, where does it dissapear after changing the variable?
    – Student123
    Nov 19 at 9:02










  • @Student123 You are welcome! The '$x$' in front of the square root came under the '$dx$' to produce $displaystyle frac{dx}{x}$ which is equal to $du$.
    – Olivier Oloa
    Nov 19 at 9:49




















up vote
3
down vote













once we preform the change of variables $u=log x$, we of course have
$$I=intfrac{mathrm{d}u}{sqrt{1+u^2}}$$
Which can be computed using the following identity with hyperbolic trig. functions:
$$cosh^2t-sinh^2t=1$$
Substitution, baby: $u=sinh tRightarrow mathrm{d}u=cosh t mathrm{d}t$
$$I=intfrac{cosh t mathrm{d}t}{sqrt{1+sinh^2t}}$$
$$I=intfrac{cosh t mathrm{d}t}{sqrt{cosh^2t}}$$
$$I=intfrac{cosh t mathrm{d}t}{cosh t}$$
$$I=intmathrm{d}t$$
$$I=t$$
$$I=text{arcsinh},u$$
$$I=text{arcsinh},log x$$
Noting that $$text{arcsinh},x=logbig(sqrt{x^2+1}+xbig)$$
Of course provides the integral:
$$I=logbigg(sqrt{log^2x+1}+log xbigg)$$
QED



Remember that $log x$ is the natural logarithm.



Edit:



adding the absolute value bars in like so:
$$I=logbigg|sqrt{log^2x+1}+log xbigg|$$
Extends the domain of the anti-derivative, which is useful if required.



Edit 2.0:



plugging in the endpoints:
$$Ibig|_1^e =logbigg(sqrt{log^2e+1}+log ebigg)-logbigg(sqrt{log^21+1}+log 1bigg)$$
$$Ibig|_1^e =logbigg(sqrt{1+1}+1bigg)-logbigg(sqrt{0+1}+0bigg)$$
$$Ibig|_1^e =logbig(sqrt{2}+1big)$$
QED






share|cite|improve this answer























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    2 Answers
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    2 Answers
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    up vote
    7
    down vote













    Hint. One may perform the change of variable
    $$
    u=ln x,qquad du=frac{dx}x,
    $$
    giving
    $$
    int frac{1}{xsqrt{1+ln^2x}}:dx=int frac{du}{sqrt{1+u^2}}
    $$
    then one may notice that
    $$
    left[ln left(u+ sqrt{1+u^2}right)right]'=frac{1+frac{u}{sqrt{1+u^2}}}{u+sqrt{1+u^2}}=frac{1}{sqrt{1+u^2}}.
    $$






    share|cite|improve this answer





















    • thank you so much for helping! but i have a question: before you change the variable there is a 'x' in front of the square root, where does it dissapear after changing the variable?
      – Student123
      Nov 19 at 9:02










    • @Student123 You are welcome! The '$x$' in front of the square root came under the '$dx$' to produce $displaystyle frac{dx}{x}$ which is equal to $du$.
      – Olivier Oloa
      Nov 19 at 9:49

















    up vote
    7
    down vote













    Hint. One may perform the change of variable
    $$
    u=ln x,qquad du=frac{dx}x,
    $$
    giving
    $$
    int frac{1}{xsqrt{1+ln^2x}}:dx=int frac{du}{sqrt{1+u^2}}
    $$
    then one may notice that
    $$
    left[ln left(u+ sqrt{1+u^2}right)right]'=frac{1+frac{u}{sqrt{1+u^2}}}{u+sqrt{1+u^2}}=frac{1}{sqrt{1+u^2}}.
    $$






    share|cite|improve this answer





















    • thank you so much for helping! but i have a question: before you change the variable there is a 'x' in front of the square root, where does it dissapear after changing the variable?
      – Student123
      Nov 19 at 9:02










    • @Student123 You are welcome! The '$x$' in front of the square root came under the '$dx$' to produce $displaystyle frac{dx}{x}$ which is equal to $du$.
      – Olivier Oloa
      Nov 19 at 9:49















    up vote
    7
    down vote










    up vote
    7
    down vote









    Hint. One may perform the change of variable
    $$
    u=ln x,qquad du=frac{dx}x,
    $$
    giving
    $$
    int frac{1}{xsqrt{1+ln^2x}}:dx=int frac{du}{sqrt{1+u^2}}
    $$
    then one may notice that
    $$
    left[ln left(u+ sqrt{1+u^2}right)right]'=frac{1+frac{u}{sqrt{1+u^2}}}{u+sqrt{1+u^2}}=frac{1}{sqrt{1+u^2}}.
    $$






    share|cite|improve this answer












    Hint. One may perform the change of variable
    $$
    u=ln x,qquad du=frac{dx}x,
    $$
    giving
    $$
    int frac{1}{xsqrt{1+ln^2x}}:dx=int frac{du}{sqrt{1+u^2}}
    $$
    then one may notice that
    $$
    left[ln left(u+ sqrt{1+u^2}right)right]'=frac{1+frac{u}{sqrt{1+u^2}}}{u+sqrt{1+u^2}}=frac{1}{sqrt{1+u^2}}.
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 18 at 0:38









    Olivier Oloa

    107k17175293




    107k17175293












    • thank you so much for helping! but i have a question: before you change the variable there is a 'x' in front of the square root, where does it dissapear after changing the variable?
      – Student123
      Nov 19 at 9:02










    • @Student123 You are welcome! The '$x$' in front of the square root came under the '$dx$' to produce $displaystyle frac{dx}{x}$ which is equal to $du$.
      – Olivier Oloa
      Nov 19 at 9:49




















    • thank you so much for helping! but i have a question: before you change the variable there is a 'x' in front of the square root, where does it dissapear after changing the variable?
      – Student123
      Nov 19 at 9:02










    • @Student123 You are welcome! The '$x$' in front of the square root came under the '$dx$' to produce $displaystyle frac{dx}{x}$ which is equal to $du$.
      – Olivier Oloa
      Nov 19 at 9:49


















    thank you so much for helping! but i have a question: before you change the variable there is a 'x' in front of the square root, where does it dissapear after changing the variable?
    – Student123
    Nov 19 at 9:02




    thank you so much for helping! but i have a question: before you change the variable there is a 'x' in front of the square root, where does it dissapear after changing the variable?
    – Student123
    Nov 19 at 9:02












    @Student123 You are welcome! The '$x$' in front of the square root came under the '$dx$' to produce $displaystyle frac{dx}{x}$ which is equal to $du$.
    – Olivier Oloa
    Nov 19 at 9:49






    @Student123 You are welcome! The '$x$' in front of the square root came under the '$dx$' to produce $displaystyle frac{dx}{x}$ which is equal to $du$.
    – Olivier Oloa
    Nov 19 at 9:49












    up vote
    3
    down vote













    once we preform the change of variables $u=log x$, we of course have
    $$I=intfrac{mathrm{d}u}{sqrt{1+u^2}}$$
    Which can be computed using the following identity with hyperbolic trig. functions:
    $$cosh^2t-sinh^2t=1$$
    Substitution, baby: $u=sinh tRightarrow mathrm{d}u=cosh t mathrm{d}t$
    $$I=intfrac{cosh t mathrm{d}t}{sqrt{1+sinh^2t}}$$
    $$I=intfrac{cosh t mathrm{d}t}{sqrt{cosh^2t}}$$
    $$I=intfrac{cosh t mathrm{d}t}{cosh t}$$
    $$I=intmathrm{d}t$$
    $$I=t$$
    $$I=text{arcsinh},u$$
    $$I=text{arcsinh},log x$$
    Noting that $$text{arcsinh},x=logbig(sqrt{x^2+1}+xbig)$$
    Of course provides the integral:
    $$I=logbigg(sqrt{log^2x+1}+log xbigg)$$
    QED



    Remember that $log x$ is the natural logarithm.



    Edit:



    adding the absolute value bars in like so:
    $$I=logbigg|sqrt{log^2x+1}+log xbigg|$$
    Extends the domain of the anti-derivative, which is useful if required.



    Edit 2.0:



    plugging in the endpoints:
    $$Ibig|_1^e =logbigg(sqrt{log^2e+1}+log ebigg)-logbigg(sqrt{log^21+1}+log 1bigg)$$
    $$Ibig|_1^e =logbigg(sqrt{1+1}+1bigg)-logbigg(sqrt{0+1}+0bigg)$$
    $$Ibig|_1^e =logbig(sqrt{2}+1big)$$
    QED






    share|cite|improve this answer



























      up vote
      3
      down vote













      once we preform the change of variables $u=log x$, we of course have
      $$I=intfrac{mathrm{d}u}{sqrt{1+u^2}}$$
      Which can be computed using the following identity with hyperbolic trig. functions:
      $$cosh^2t-sinh^2t=1$$
      Substitution, baby: $u=sinh tRightarrow mathrm{d}u=cosh t mathrm{d}t$
      $$I=intfrac{cosh t mathrm{d}t}{sqrt{1+sinh^2t}}$$
      $$I=intfrac{cosh t mathrm{d}t}{sqrt{cosh^2t}}$$
      $$I=intfrac{cosh t mathrm{d}t}{cosh t}$$
      $$I=intmathrm{d}t$$
      $$I=t$$
      $$I=text{arcsinh},u$$
      $$I=text{arcsinh},log x$$
      Noting that $$text{arcsinh},x=logbig(sqrt{x^2+1}+xbig)$$
      Of course provides the integral:
      $$I=logbigg(sqrt{log^2x+1}+log xbigg)$$
      QED



      Remember that $log x$ is the natural logarithm.



      Edit:



      adding the absolute value bars in like so:
      $$I=logbigg|sqrt{log^2x+1}+log xbigg|$$
      Extends the domain of the anti-derivative, which is useful if required.



      Edit 2.0:



      plugging in the endpoints:
      $$Ibig|_1^e =logbigg(sqrt{log^2e+1}+log ebigg)-logbigg(sqrt{log^21+1}+log 1bigg)$$
      $$Ibig|_1^e =logbigg(sqrt{1+1}+1bigg)-logbigg(sqrt{0+1}+0bigg)$$
      $$Ibig|_1^e =logbig(sqrt{2}+1big)$$
      QED






      share|cite|improve this answer

























        up vote
        3
        down vote










        up vote
        3
        down vote









        once we preform the change of variables $u=log x$, we of course have
        $$I=intfrac{mathrm{d}u}{sqrt{1+u^2}}$$
        Which can be computed using the following identity with hyperbolic trig. functions:
        $$cosh^2t-sinh^2t=1$$
        Substitution, baby: $u=sinh tRightarrow mathrm{d}u=cosh t mathrm{d}t$
        $$I=intfrac{cosh t mathrm{d}t}{sqrt{1+sinh^2t}}$$
        $$I=intfrac{cosh t mathrm{d}t}{sqrt{cosh^2t}}$$
        $$I=intfrac{cosh t mathrm{d}t}{cosh t}$$
        $$I=intmathrm{d}t$$
        $$I=t$$
        $$I=text{arcsinh},u$$
        $$I=text{arcsinh},log x$$
        Noting that $$text{arcsinh},x=logbig(sqrt{x^2+1}+xbig)$$
        Of course provides the integral:
        $$I=logbigg(sqrt{log^2x+1}+log xbigg)$$
        QED



        Remember that $log x$ is the natural logarithm.



        Edit:



        adding the absolute value bars in like so:
        $$I=logbigg|sqrt{log^2x+1}+log xbigg|$$
        Extends the domain of the anti-derivative, which is useful if required.



        Edit 2.0:



        plugging in the endpoints:
        $$Ibig|_1^e =logbigg(sqrt{log^2e+1}+log ebigg)-logbigg(sqrt{log^21+1}+log 1bigg)$$
        $$Ibig|_1^e =logbigg(sqrt{1+1}+1bigg)-logbigg(sqrt{0+1}+0bigg)$$
        $$Ibig|_1^e =logbig(sqrt{2}+1big)$$
        QED






        share|cite|improve this answer














        once we preform the change of variables $u=log x$, we of course have
        $$I=intfrac{mathrm{d}u}{sqrt{1+u^2}}$$
        Which can be computed using the following identity with hyperbolic trig. functions:
        $$cosh^2t-sinh^2t=1$$
        Substitution, baby: $u=sinh tRightarrow mathrm{d}u=cosh t mathrm{d}t$
        $$I=intfrac{cosh t mathrm{d}t}{sqrt{1+sinh^2t}}$$
        $$I=intfrac{cosh t mathrm{d}t}{sqrt{cosh^2t}}$$
        $$I=intfrac{cosh t mathrm{d}t}{cosh t}$$
        $$I=intmathrm{d}t$$
        $$I=t$$
        $$I=text{arcsinh},u$$
        $$I=text{arcsinh},log x$$
        Noting that $$text{arcsinh},x=logbig(sqrt{x^2+1}+xbig)$$
        Of course provides the integral:
        $$I=logbigg(sqrt{log^2x+1}+log xbigg)$$
        QED



        Remember that $log x$ is the natural logarithm.



        Edit:



        adding the absolute value bars in like so:
        $$I=logbigg|sqrt{log^2x+1}+log xbigg|$$
        Extends the domain of the anti-derivative, which is useful if required.



        Edit 2.0:



        plugging in the endpoints:
        $$Ibig|_1^e =logbigg(sqrt{log^2e+1}+log ebigg)-logbigg(sqrt{log^21+1}+log 1bigg)$$
        $$Ibig|_1^e =logbigg(sqrt{1+1}+1bigg)-logbigg(sqrt{0+1}+0bigg)$$
        $$Ibig|_1^e =logbig(sqrt{2}+1big)$$
        QED







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered Nov 18 at 6:42









        clathratus

        2,111220




        2,111220






























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