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Math Help - another definite integral

  1. #1
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    another definite integral

    I am trying to evaluate \int_{0}^{1}\frac{\tan^{-1}{x}}{x+1}dx.

    I used integration by parts and got

    \frac{\pi}{4} \ln(2) - \int_{0}^{1} \frac{\ln(x+1)}{x^2+1}dx

    How do I solve the second integral?

    thanks!
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  2. #2
    Member sbhatnagar's Avatar
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    Lightbulb Re: another definite integral

    Let I=\int_{0}^{1} \frac{\ln(x+1)}{x^2+1}dx.

    1. Substitute x = \tan{u} and dx=\sec^2{u} \ du.

    I=\int_{0}^{\pi \over 4} \frac{\sec^2{u} \cdot \ln(\tan{u}+1)}{\sec^2{u}}dx

    \implies I=\int_{0}^{\pi \over 4}  \ln(\tan{u}+1) \ dx

    2. Use the identity \int_{0}^{a}f(x) \ dx = \int_{0}^{a}f(a-x) \ dx:

    I=\int_{0}^{\pi \over 4}  \ln\left(\tan{\left( \frac{\pi}{4}- u \right) }+1 \right) \ dx

    3. Use the identity \tan(a-b)=\frac{\tan{a}-\tan{b}}{1+\tan{a}\tan{b}}:

    I=\int_{0}^{\pi \over 4}  \ln\left( \frac{2}{1+\tan{u}}\right) \ dx

    \implies I=\int_{0}^{\pi \over 4}  \ln(2) \ dx - \int_{0}^{\pi \over 4} \ln(1+\tan{u}) \ dx

    \implies I= \int_{0}^{\pi \over 4}  \ln(2) \ dx -I

    \implies 2I=\frac{\pi}{4}\ln(2)

    \implies I=\frac{\pi}{8}\ln(2)
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  3. #3
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    Re: another definite integral

    Quote Originally Posted by sbhatnagar View Post
    Let I=\int_{0}^{1} \frac{\ln(x+1)}{x^2+1}dx.

    1. Substitute x = \tan{u} and dx=\sec^2{u} \ du.

    I=\int_{0}^{\pi \over 4} \frac{\sec^2{u} \cdot \ln(\tan{u}+1)}{\sec^2{u}}dx

    \implies I=\int_{0}^{\pi \over 4}  \ln(\tan{u}+1) \ dx

    2. Use the identity \int_{0}^{a}f(x) \ dx = \int_{0}^{a}f(a-x) \ dx:

    I=\int_{0}^{\pi \over 4}  \ln\left(\tan{\left( \frac{\pi}{4}- u \right) }+1 \right) \ dx

    3. Use the identity \tan(a-b)=\frac{\tan{a}-\tan{b}}{1+\tan{a}\tan{b}}:

    I=\int_{0}^{\pi \over 4}  \ln\left( \frac{2}{1+\tan{u}}\right) \ dx

    \implies I=\int_{0}^{\pi \over 4}  \ln(2) \ dx - \int_{0}^{\pi \over 4} \ln(1+\tan{u}) \ dx

    \implies I= \int_{0}^{\pi \over 4}  \ln(2) \ dx -I

    \implies 2I=\frac{\pi}{4}\ln(2)

    \implies I=\frac{\pi}{8}\ln(2)
    woa! thanks!! You used a very tricky approach.

    That should mean : \int_{0}^{1}\frac{\tan^{-1}{x}}{1+x}dx=\frac{\pi}{4}\ln(2)-\frac{\pi}{8}\ln(2)=\frac{\pi}{8}\ln(2)
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