Question

If $I=\int \frac{dx}{x\sqrt{x^4+1}}$, then I equals

• A. $\left(1/4\right)\log \left|\frac{\sqrt{x^4+1}-1}{\sqrt{x^4+1}+1}\right|+C$
• B. $\left(1/4\right)\log \left|\frac{\sqrt{x^4+1}+1}{\sqrt{x^4+1}-1}\right|+C$
• C. $\left(1/4\right)(x^4+1{)}^{3/2}+C$
• D. $\left(1/4\right)(x^4-1{)}^{-3/2}+C$

Answer 1

The correct option is A $\left(1/4\right)\log \left|\frac{\sqrt{x^4+1}-1}{\sqrt{x^4+1}+1}\right|+C$

$I=\int \frac{dx}{x\sqrt{x^4+1}}$
Substitute $x^4+1=t^2$
$I=\frac{1}{2}\int \frac{tdt}{(t^2-1)t}=\frac{1}{4}\log \left|\frac{t-1}{t+1}\right|+c=\frac{1}{4}\log \left|\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+1}+1}\right|+c$