Evaluate $\int \frac{\sqrt{1 + x^2}}{x^4} \, dx$.

$-\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{3/2}+C$

The correct answer is Option (3) → $-\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{3/2}+C$

Let $I = \int \frac{\sqrt{1 + x^2}}{x^4} \, dx = \int \frac{\sqrt{1 + x^2}}{x} \cdot \frac{1}{x^3} \, dx \text{}$

$= \int \sqrt{\frac{1 + x^2}{x^2}} \cdot \frac{1}{x^3} \, dx = \int \sqrt{\frac{1}{x^2} + 1} \cdot \frac{1}{x^3} \, dx \text{}$

Put $1 + \frac{1}{x^2} = t^2 ⇒\frac{-2}{x^3} \, dx = 2t \, dt$

$⇒-\frac{1}{x^3}.dx=t.dt$

$∴I=-\int t^2dt=-\frac{t^3}{3}+C=-\frac{1}{3}\left(1+\frac{1}{x^2}\right)^{3/2}+C$