alleryStart learning →$\displaystyle\int_0^1 \dfrac{\ln(1+x)}{1+x^2}\, dx = ?$ (Putnam classic)
- $(\pi/8)\ln 2$
- $(\pi/4)\ln 2$
- $\ln 2$
- $\pi/8$
Show answer and 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 $2I = (\pi/4)\ln 2$ ⇒ $I = (\pi/8)\ln 2$.**