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

#### 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.

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.

#### Plato

MHF Helper
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

#### Idea

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

$-0.915966$

1 person

Oh, thank you!