Question

If $ \int \frac{\sqrt{1-x^{2}}}{x^{4}} dx = A \left(x\right)\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 :

• A. $\frac{-1}{3x^3}$
• B. $\frac{-1}{27 x^9}$
• C. $\frac{1}{9 x^4}$
• D. $\frac{1}{27 x^6}$

Answer 1

Answer: (B)

$\int \frac{\sqrt{1-x^{2}}}{x^{4}} dx = A\left(x\right) \left(\sqrt{1-x^{2}}\right)^{m} +C $
$ \int \frac{\left|x \right|\sqrt{\frac{1}{x^{2}} -1}}{x^{4}} dx $
Put $ \frac{1}{x^{2} } - 1 = t \Rightarrow \frac{dt}{dx} = \frac{-2}{x^{3}}$
Case-1 x $\ge$ 0
$ - \frac{1}{2} \int \sqrt{t} dt \Rightarrow - \frac{t^{3/2}}{3} + C $
$ \Rightarrow - \frac{1}{3} \left(\frac{1}{x^{2}}-1\right) ^{3/2} $
$ \Rightarrow \frac{\left(\sqrt{1-x^{2}}\right)^{3}}{-3x^{2}} +C $
$ A\left(x\right) = - \frac{1}{3x^{3}} $
$ \left(A\left(x\right)\right)^{m} = \left(- \frac{1}{3x^{3}}\right)^{3} = - \frac{1}{27x^{9}} $
Case-II x $\le$ 0
We get $\frac{\left(\sqrt{1-x^{2}}\right)^{3}}{-3x^{3}} +C$
$ A\left(x\right) = \frac{1}{-3x^{3}} , m = 3 $
$ \left(A\left(x\right)\right)^{m} = \frac{-1}{27x^{9}} $