Topic: Engineering Mathematics
Evaluate $\int_{0}^{1}\frac{\ln(x+1)}{x^{2}+1}\cdot dx$
- A. $\pi\ln\sqrt{2}$
- B. $\frac{\pi}{8}\ln 2$
- C. $2\pi\ln\sqrt{2}$
- D. $\ln\sqrt{2}$
Answer: B. $\frac{\pi}{8}\ln 2$
Explanation: Let $x=\tan y$, then $dx=\sec^{2}y dy$.
$\int_{0}^{1}\frac{\ln(x+1)}{x^{2}+1}dx = \int_{0}^{\pi/4}\frac{\ln(1+\tan y)}{\sec^{2}y}\sec^{2}y dy = \int_{0}^{\pi/4}\ln(1+\tan y)dy = I$.
Using the property $\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$:
$I = \int_{0}^{\pi/4}\ln(1+\tan(\pi/4-y))dy = \int_{0}^{\pi/4}\ln(1+\frac{1-\tan y}{1+\tan y})dy = \int_{0}^{\pi/4}\ln(\frac{2}{1+\tan y})dy$.
Adding both expressions for I:
$2I = \int_{0}^{\pi/4}\ln(2)dy \Rightarrow I = \frac{\pi}{8}\ln(2)$.