allery$\displaystyle\int_0^1 \dfrac{\ln(1+x)}{1+x^2}\, dx = ?$ (Putnam classic)
A$(\pi/8)\ln 2$
B$(\pi/4)\ln 2$
C$\ln 2$
D$\pi/8$
Answer & Solution
Correct answer: A. $(\pi/8)\ln 2$
Sub $x = \tan\theta$, $dx = \sec^2\theta d\theta$: $I = \int_0^{\pi/4} \ln(1 + \tan\theta)\, d\theta$.
**King's:** $\theta \to \pi/4 - \theta$: $\tan(\pi/4 - \theta) = (1 - \tan\theta)/(1 + \tan\theta)$
so $1 + \tan(\pi/4 - \theta) = 2/(1 + \tan\theta)$.
Thus $I = \int_0^{\pi/4}[\ln 2 - \ln(1 + \tan\theta)]\, d\theta = (\pi/4)\ln 2 - I$.
**So $2I = (\pi/4)\ln 2$ ⇒ $I = (\pi/8)\ln 2$.**
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