Question

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

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