How do i calculate the value of $\int_{0}^{1} \frac{\ln{(1+x)}}{1+x^2}$?

Question

How do i calculate the value of the following integral--

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

I tried doing substitutions like $1+x=t$ etc.

I also tried to use the property $\displaystyle\int_{a}^{b} f(x)=\displaystyle\int_{a}^{b} f(a+b-x)$, but didn't arrive at anything fruitful.

Doing integration by parts after the sub $1 + x = t$ leads to the integral of $ln(t)/t-2$ which involves the polylog function (so that is not the way to go). — AnonymousCoward, Jul 23 '14 at 16:13
http://math.stackexchange.com/questions/220746/integrating-frac-log1x1x2 — lab bhattacharjee, Jul 23 '14 at 16:17

evinda · Accepted Answer · 2014-07-23T16:20:18.227

Use the substitution: $x = \tan\theta$.

The integral is then equal to: $$I= \int_{0}^{\pi/4} \ln(1+\tan\theta) \ d\theta (*)$$

Also,we know the property: $$\int_{0}^{b} f(x) \ dx = \int_{0}^{b} f(b-x) \ dx$$ so we have $$I = \int_{0}^{\pi/4} \ln\biggl(1+\tan\Bigl(\frac{\pi}{4}-\theta\Bigr)\biggr) \ d\theta = \int_{0}^{\pi/4} \ln\biggl(\frac{2}{1+\tan\theta} \biggr) \ d\theta (**)$$

$(*)+(**) \Rightarrow 2I = \int_{0}^{\pi/4} \ln 2 \ d\theta\Rightarrow I= \ln 2 \cdot \frac{\pi}{8}$

score 0 · Answer 2 · answered Jul 23 '14 at 16:09

0

Try using $x=\tan\theta$ for the substitution. Then use the second result you have mentioned in your question with new limits. You may want to use the formula for $\tan (A+B)$ at some point.

answered Jul 23 '14 at 16:09

Anurag A

41,067

How do i calculate the value of $\int_{0}^{1} \frac{\ln{(1+x)}}{1+x^2}$?

2 Answers2