# If the solve the problem

Question:

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

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

1. $\frac{-1}{3 x^{3}}$

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

3. $\frac{1}{9 x^{4}}$

4. $\frac{1}{27 x^{6}}$

Correct Option: , 2

Solution:

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

$\int \frac{|x| \sqrt{\frac{1}{x^{2}}-1}}{x^{4}} d x$

Put $\frac{1}{\mathrm{x}^{2}}-1=\mathrm{t} \Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{-2}{\mathrm{x}^{3}}$

Case-1 $x \geq 0$

$-\frac{1}{2} \int \sqrt{t} \mathrm{dt} \Rightarrow-\frac{\mathrm{t}^{3 / 2}}{3}+\mathrm{C}$

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

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

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

$(\mathrm{A}(\mathrm{x}))^{\mathrm{m}}=\left(-\frac{1}{3 \mathrm{x}^{3}}\right)^{3}=-\frac{1}{27 \mathrm{x}^{9}}$

Case-II $\mathrm{x} \leq 0$

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

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

$(A(x))^{m}=\frac{-1}{27 x^{9}}$