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

Answer 1

Answer: (B)

$\int \frac{\sqrt{1-x^2}}{x^4}dx=A\left(x\right){\left(\sqrt{1-x^2}\right)}^m+C$
$\int \frac{|x|\sqrt{\frac{1}{x^2}-1}}{x^4}dx$
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 $\frac{{\left(\sqrt{1-x^2}\right)}^3}{-3x^3}+C$
$A\left(x\right)=\frac{1}{-3x^3},m=3$
${\left(A\left(x\right)\right)}^m=\frac{-1}{27x^9}$