$\int \frac{dx}{x^{3}(1+x^{4})^{\frac{1}{2}}}$ equals:

$-\frac{1}{2x^{2}} \sqrt{1+x^{4}} + c$

The correct answer is Option (1) → $-\frac{1}{2x^{2}} \sqrt{1+x^{4}} + c$

Let $I = \int \frac{dx}{x^{3}(1+x^{4})^{\frac{1}{2}}} = \int \frac{dx}{x^{5} \left( 1 + \frac{1}{x^{4}} \right)^{\frac{1}{2}}}$

$\text{(Let } 1 + x^{-4} = 1 + \frac{1}{x^{4}} = t,  dt = -4x^{-5} dx ⇒\frac{dx}{x^{5}} = -\frac{1}{4} dt \text{)}$

$= -\frac{1}{4} \int \frac{dt}{t^{1/2}} = -\frac{1}{4} \times 2 \times \sqrt{t} + c$, where 'c' denotes any arbitrary constant of integration.

$= -\frac{1}{2} \sqrt{1 + \frac{1}{x^4}} + c$

$= -\frac{1}{2} \frac{\sqrt{x^4+1}}{x^2} + c$