Question

$\int \frac{x\cdot \log x}{{\left(\sqrt{x^2-1}\right)}^3}dx=$

• A. ${\sec }^{-1}x+\frac{\log x}{\sqrt{x^2-1}}+C$
• B. ${\sec }^{-1}x-\frac{\log x}{\sqrt{x^2-1}}+C$
• C. $\frac{\log x}{\sqrt{x^2-1}}-{\sec }^{-1}x+C$
• D. $\frac{-\log x}{\sqrt{x^2-1}}-{\sec }^{-1}x+C$

Answer 1

Answer: (B)

Let
$I=\int \frac{x\log x}{(\sqrt{{x^2-1)}^3}}dx$
Let $\sqrt{x^2-1}=t\Rightarrow \frac{2xdx}{2\sqrt{x^2-1}}=dt\Rightarrow \frac{xdx}{\sqrt{x^2-1}}=dt$
$\therefore I=\int \frac{\log xdt}{t^2}$
$=\int \frac{\log \left(\sqrt{1+t^2}\right)}{t^2}dt=\int \log \sqrt{\left(1+t^2\right)}\cdot \left(\frac{1}{t^2}\right)dt$
$\left[\because \sqrt{x^2-1}=t\right]\Rightarrow x^2=t^2+1\Rightarrow x=\sqrt{1+t^2}]$
$=-\frac{1}{t}\log \sqrt{1+t^2}-\int \frac{2t}{2\sqrt{1+t^2}}\cdot \left(-\frac{1}{t}\right)\cdot \frac{1}{\sqrt{1+t^2}}dt$
$=-\frac{1}{t}\log \sqrt{1+t^2}+\int \frac{dt}{1+t^2}$
$=-\frac{1}{t}\log \sqrt{1+t^2}+{\tan }^{-1}(t)+c$
${\tan }^{-1}\left(\sqrt{x^2-1}\right)-\frac{\log x}{\sqrt{x^2-1}}+c$
$={\sec }^{-1}(x)-\frac{\log x}{\sqrt{x^2-1}}+c$