1. ## 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!

2. ## 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 $\tan(a-b)=\frac{\tan{a}-\tan{b}}{1+\tan{a}\tan{b}}$:

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

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

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

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

$\implies I=\frac{\pi}{8}\ln(2)$

3. ## Re: another definite integral

Originally Posted by sbhatnagar
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 : $\int_{0}^{1}\frac{\tan^{-1}{x}}{1+x}dx=\frac{\pi}{4}\ln(2)-\frac{\pi}{8}\ln(2)=\frac{\pi}{8}\ln(2)$