Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question already has an answer here:

How can I calculate the integral $I=\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx$ by substituting $t=\frac{1-x}{1+x}$.

share|cite|improve this question

marked as duplicate by gebruiker, Watson, G. Sassatelli, Jennifer, RecklessReckoner Jun 3 at 19:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

@RonGordon That's why I put here because it's not an ordinary integration by substitution. – pourjour May 6 '13 at 22:37
@pourjour You mean that you're actually integrating from $0$ to $\frac{1-x}{1+x}$? (I don't see how that makes sense) – apnorton May 6 '13 at 22:40
@anorton sorry I meant 1 not t – pourjour May 6 '13 at 22:44
up vote 10 down vote accepted

OK, the substitution you suggest works very well. Note that

$$x = \frac{1-t}{1+t}$$

$$1+x = \frac{2}{1+t}$$

$$1+x^2 = \frac{2(1+t^2)}{(1+t)^2}$$

$$dx = -\frac{2}{(1+t)^2} dt$$


$$\int_0^1 dx \frac{\log{(1+x)}}{1+x^2} = \int_0^1 dt \frac{\log{2} - \log{(1+t)}}{1+t^2}$$

With a little bit of algebra, we see immediately that

$$\int_0^1 dx \frac{\log{(1+x)}}{1+x^2} = \frac{\pi}{8} \log{2}$$

share|cite|improve this answer

Another way. Setting $x= \tan(t)$, we get that \begin{align} I & = \int_0^{\pi/4} \ln(1+\tan(t))dt = \int_0^{\pi/4} \ln \left(\sqrt2 \cos(\pi/4-t)\right) dt - \int_0^{\pi/4} \ln(\cos(t)) dt\\ & = \int_0^{\pi/4} \ln(\sqrt2) dt + \underbrace{\int_0^{\pi/4}\ln \left(\cos(\pi/4-t)\right) dt}_{\pi/4-t \to t \text{ gives }\int_0^{\pi/4} \ln(\cos(t)) dt} - \int_0^{\pi/4} \ln(\cos(t)) dt\\ & = \dfrac{\pi \ln(2)}8 \end{align}

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.