Question

If $I=\int x\sqrt{\frac{x^2+1}{x^2-1}}dx,$ then $I$ equals

• A. $\frac{1}{2}\sqrt{x^4-1}+\frac{1}{2}\sqrt{x^4+1}+C$
• B. $\frac{1}{2}\sqrt{x^4-1}+\frac{1}{2}\log (x^2+\sqrt{x^4-1})+C$
• C. $\sqrt{x^4-1}+{\sin }^{-1}\left(x^2\right)+C$
• D. $\sqrt{x^4-1}+2{\sin }^{-1}\left(x^2\right)+C$

Answer 1

The correct option is B $\frac{1}{2}\sqrt{x^4-1}+\frac{1}{2}\log (x^2+\sqrt{x^4-1})+C$

Let $I=\int x\sqrt{\frac{x^2+1}{x^2-1}}dx$
Substitute $x^2=t\Rightarrow 2xdx=2t$
$\therefore I=\frac{1}{2}\int \sqrt{\frac{t+1}{t-1}}dt=\frac{1}{2}\int \frac{t+1}{\sqrt{t^2-1}}dt$
$=\frac{1}{2}\int \frac{t}{\sqrt{t^2-1}}dt+\frac{1}{2}\int \frac{dt}{\sqrt{t^2-1}}$
$=\frac{1}{2}\sqrt{t^2-1}+\frac{1}{2}\log |t+\sqrt{t^2-1}|+C$
$=\frac{1}{2}\sqrt{x^4-1}+\frac{1}{2}\log |x^2+\sqrt{x^4-1}|+C$