# Thread: How to solve improper integral ln(x) / (x^2 + 1)

1. ## How to solve improper integral ln(x) / (x^2 + 1)

Hello, guys, I just want you to help me of finding the integral of $\displaystyle \int_{0}^{1}\dfrac{\ln(x)}{x^2 + 1}\mathrm{d}x$. I just know that is improper.

2. ## Re: How to solve improper integral ln(x) / (x^2 + 1)

Use the method of integration by parts. As above, let u = ln(x), dv = 1/(x^2+1). Let me know if you need more help.

3. ## Re: How to solve improper integral ln(x) / (x^2 + 1)

Originally Posted by lebdim
Hello, guys, I just want you to help me of finding the integral of $\displaystyle \int_{0}^{1}\dfrac{\ln(x)}{x^2 + 1}\mathrm{d}x$. I just know that is improper.
Is your difficulty with the anti-derivative $\displaystyle \displaystyle\int {\frac{{\log (x)}}{{{x^2} + 1}}dx}~?$
There is no elementary solution. The solution involves the Polylogarithm function: SEE HERE

4. ## Re: How to solve improper integral ln(x) / (x^2 + 1)

integration by parts we get

$\displaystyle -\int_0^1 \frac{\tan ^{-1}x}{x} \, dx$

$\displaystyle -\int_0^1 \left(\sum _{k=0}^{\infty } \frac{(-1)^kx^{2k}}{(2k+1)}\right) \, dx$

$\displaystyle -\sum _{k=0}^{\infty } \frac{(-1)^k}{(2k+1)}\int_0^1 x^{2k} \, dx$

$\displaystyle -\sum _{k=0}^{\infty } \frac{(-1)^k}{(2k+1)^2}$

$\displaystyle -Catalan$ constant

$-0.915966$

5. ## Re: How to solve improper integral ln(x) / (x^2 + 1)

Oh, thank you!