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I'm trying to evaluate the definite integral $I = \displaystyle\int_0^1 \frac{\ln(1-x)}{1+x}dx$.

My first idea was integration by parts. I set $u = \ln(1-x)$ and $dv = \frac{1}{1+x}dx$. This gives $du = \frac{-1}{1-x}dx$ and $v = \ln(1+x)$, leading to the new integral: $$ \int_0^1 \frac{\ln(1+x)}{1-x}dx $$ This doesn't seem any simpler than the original problem.

Is there another method or a specific substitution I should be trying?

navrug
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  • Your integration by parts isn’t quite right. You can check that your new integral diverges. The antiderivative of $\frac{1}{1+x}$ should really be $\ln(1+x)+c$ where $c$ is some arbitrary constant. Can you find a value for $c$ that makes the integral converge? – David H Jun 29 '25 at 14:30
  • BTW, Integral calculator gives the answer $\frac{\ln^22}{2}-\frac{\pi^2}{12}$ by finding the antiderivative $(\ln 2)\ln(\vert x+1\vert)-\text{Li}_2\left(\frac{x+1}{2}\right)+C$. You may check the steps if you are interested. – Bowei Tang Jun 29 '25 at 14:32
  • I feel like the method of differentiating under the integral sign could be helpful here, but I am not certain. – Tom Jun 29 '25 at 14:36

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