chosen integer $m$ and a function A(x)
Question:

If $\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x=A(\mathrm{x})\left(\sqrt{1-x^{2}}\right)^{m}+C$, for a suitable

chosen integer $m$ and a function $A(x)$, where $C$ is a constant of integration, then $(\mathrm{A}(\mathrm{x}))^{\mathrm{m}}$ equals :

1. (1) $\frac{-1}{27 x^{9}}$

2. (2) $\frac{-1}{3 x^{3}}$

3. (3) $\frac{1}{27 x^{6}}$

4. (4) $\frac{1}{9 x^{4}}$

Correct Option: 1

Solution:

$A(x)\left(\sqrt{1-x^{2}}\right)^{m}+C=\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x$

$=\int \frac{\sqrt{\frac{1}{x^{2}}-1}}{x^{3}} d x$

Let $\frac{1}{x^{2}}-1=u^{2}$

$\Rightarrow \quad-\frac{2}{x^{3}}=\frac{2 u d u}{d x}$

$\frac{d x}{x^{3}}=-u d u$

$A(x)\left(\sqrt{1-x^{2}}\right)^{m}+C=\int\left(-u^{2}\right) d u=-\frac{u^{3}}{3}+C$

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

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

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

Compare both sides,

$\Rightarrow \quad A(x)=-\frac{1}{3 x^{3}}$ and $m=3$

$\Rightarrow \quad(A(x))^{3}=\frac{-1}{27 x^{9}}$