$\int\frac{dx}{x^3\sqrt{(1+x^4)}}=$

$-\frac{1}{2x^2}\sqrt{1+x^4}+C$, C is an arbitary constant

The correct answer is Option (4) → $-\frac{1}{2x^2}\sqrt{1+x^4}+C$, C is an arbitary constant

Given integral:

$\displaystyle \int \frac{dx}{x^{3}\sqrt{1+x^{4}}}$

Let $x^{2}=t$ ⇒ $2x\,dx=dt$ ⇒ $dx=\frac{dt}{2x}$

Substitute into the integral:

$\displaystyle \int \frac{dx}{x^{3}\sqrt{1+x^{4}}} = \int \frac{1}{x^{3}\sqrt{1+t^{2}}} \cdot \frac{dt}{2x} = \frac{1}{2}\int \frac{dt}{x^{4}\sqrt{1+t^{2}}}$

But $x^{2}=t \Rightarrow x^{4}=t^{2}$, so

$\displaystyle \frac{1}{2}\int \frac{dt}{t^{2}\sqrt{1+t^{2}}}$

Let $t=\tan\theta$ ⇒ $dt=\sec^{2}\theta\,d\theta$, $\sqrt{1+t^{2}}=\sec\theta$

$\displaystyle \frac{1}{2}\int \frac{\sec^{2}\theta\,d\theta}{\tan^{2}\theta\,\sec\theta} = \frac{1}{2}\int \frac{\sec\theta}{\tan^{2}\theta}\,d\theta = \frac{1}{2}\int \frac{\cos\theta}{\sin^{2}\theta}\,d\theta = \frac{1}{2}\int \cot\theta\,\csc\theta\,d\theta$

$= -\frac{1}{2}\csc\theta + C$

Back substitute: $\tan\theta = t = x^{2}$, so $\csc\theta = \frac{\sqrt{1+\tan^{2}\theta}}{\tan\theta} = \frac{\sqrt{1+x^{4}}}{x^{2}}$

Therefore:

$\displaystyle \int \frac{dx}{x^{3}\sqrt{1+x^{4}}} = -\frac{\sqrt{1+x^{4}}}{2x^{2}} + C$