NIMCET 2022 — Mathematics PYQ

To solve this, we rewrite the integral by factoring $x^4$ out from the square root:

$$I=\int \frac{(x^2-1)dx}{x^3\sqrt{x^4(2-\frac{2}{x^2}+\frac{1}{x^4})}}$$

$$I=\int \frac{(x^2-1)dx}{x^3\cdot x^2\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}$$

$$I=\int \frac{(x^2-1)dx}{x^5\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}$$

Now, divide the numerator by $x^5$:

$$I=\int \frac{(\frac{1}{x^3}-\frac{1}{x^5})dx}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}$$

Substitution:

Let $t=2-\frac{2}{x^2}+\frac{1}{x^4}$.

Differentiating with respect to $x$:

$$\frac{dt}{dx}=0-2(-2x^{-3})+(-4x^{-5})$$

$$\frac{dt}{dx}=\frac{4}{x^3}-\frac{4}{x^5}=4\left(\frac{1}{x^3}-\frac{1}{x^5}\right)$$

Therefore, $\left(\frac{1}{x^3}-\frac{1}{x^5}\right)dx=\frac{dt}{4}$.

Substitution into Integral:

$$I=\int \frac{1}{\sqrt{t}}\cdot \frac{dt}{4}=\frac{1}{4}\int t^{-1/2}dt$$

$$I=\frac{1}{4}\cdot \frac{t^{1/2}}{1/2}+C=\frac{1}{2}\sqrt{t}+C$$

Substituting $t$ back:

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

The correct option is C.