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

May 2012
92
5
slovenia
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.
 
Jan 2009
463
138


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
Aug 2006
22,507
8,664
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
 
Jun 2013
1,151
614
Lebanon
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

$-0.915966$
 
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