Question

The value of the integral $\underset{1}{\overset{\frac{1+\sqrt{5}}{2}}{\int }}\frac{x^2+1}{x^4-x^2+1}\ln (1+x-\frac{1}{x})dx$ is
(correct answer + 2, wrong answer - 0.50)

• A. $\frac{\pi }{8}\ln 2$
• B. $\frac{\pi }{4}\ln 2$
• C. $\frac{\pi }{2}\ln 2$
• D. $\frac{2\pi }{3}\ln 2$

Answer 1

The correct option is A $\frac{\pi }{8}\ln 2$

Given : $I=\underset{1}{\overset{\frac{1+\sqrt{5}}{2}}{\int }}\frac{x^2+1}{x^4-x^2+1}\ln (1+x-\frac{1}{x})dx$
$\Rightarrow I=\underset{1}{\overset{\frac{1+\sqrt{5}}{2}}{\int }}\frac{1+\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}}\ln (1+x-\frac{1}{x})dx$

Assuming $x-\frac{1}{x}=t\Rightarrow (1+\frac{1}{x^2})dx=dt{x}^2+\frac{1}{x^2}-2=t^2$

When $x=1\Rightarrow t=0,x=\frac{\sqrt{5}+1}{2}\Rightarrow t=1$

$\Rightarrow I=\underset{0}{\overset{1}{\int }}\frac{\ln (1+t)}{t^2+1}dt$
Assuming $t=\tan \theta \Rightarrow {\sec }^2\theta d\theta =dt$
$\Rightarrow I=\underset{0}{\overset{\pi /4}{\int }}\ln (1+\tan \theta )d\theta \cdots \left(1\right)\Rightarrow I=\underset{0}{\overset{\pi /4}{\int }}\ln (1+\tan (\frac{\pi }{4}-\theta ))d\theta \Rightarrow I=\underset{0}{\overset{\pi /4}{\int }}\ln \left(\frac{2}{1+\tan \theta }\right)d\theta \cdots \left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$, we get
$\Rightarrow 2I=\frac{\pi }{4}\ln 2\therefore I=\frac{\pi }{8}\ln 2$