Putnam 2005 
Topic: Calculus

Problem: Evaluate $\int_0^1 \frac{\ln(x+1)}{x^2+1}\,dx$.

Solution: We make the substitution $x = \tan \theta$, rewriting the desired integral as \[ \int_0^{\pi/4} \log(\tan(\theta) + 1)\,d\theta. \] Write \[ \log(\tan(\theta)+ 1) = \log(\sin(\theta) + \cos(\theta))-\log(\cos(\theta)) \] and then note that $\sin(\theta) + \cos(\theta) = \sqrt{2} \cos (\pi/4 - \theta)$. We may thus rewrite the integrand as \[ \frac12 \log(2) + \log(\cos(\pi/4 - \theta)) - \log(\cos(\theta)). \] But over the interval $[0, \pi/4]$, the integrals of $\log(\cos(\theta))$ and $\log(\cos(\pi/4 - \theta))$ are equal, so their contributions cancel out. The desired integral is then just the integral of $\frac{1}{2} \log(2)$ over the interval $[0,\pi/4]$, which is $\boxed{\pi \log(2)/8}$.