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 $A\left(x\right)$, where $C$ is a constant of integration, then ${\left(A\right(x\left)\right)}^m$ equals :

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

Answer 1

The correct option is C $\frac{-1}{27x^9}$

Let $I=\int \frac{\sqrt{1-x^2}}{x^4}dx$

$I=\int \frac{|x|\sqrt{\frac{1}{x^2}-1}}{x^4}dx$

$\underset{–}{\text{Case I: If}x>0}$

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

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

$\Rightarrow I=\int \frac{\sqrt{t}}{-2}dt=\frac{t^{3/2}}{\frac{3}{2}\times (-2)}=-\frac{t^{3/2}}{3}$

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

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

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

$\underset{–}{\text{Case II: If}x<0}$

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

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

$\Rightarrow I=\int \frac{\sqrt{t}}{2}dt=\frac{t^{3/2}}{\frac{3}{2}\times 2}=\frac{t^{3/2}}{3}$

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

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

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