Question

If $\int \frac{\sqrt{1-x^2}}{x^4}dx=A\left(x\right)\left(\sqrt{1-x^2}\right)+C$, for a suitable chosen integer m and a function
A(x), where C is a constant of integration then $\left(A\right(x){)}^m$ equals :

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

Answer 1

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

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

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

Put $\frac{1}{x^2}-1=t\Rightarrow \frac{dt}{dx}=\frac{-2}{x^3}$
Case $-1$ $x\ge 0$
$-\frac{1}{2}\int \sqrt{t}dt\Rightarrow -\frac{t^{\frac{3}{2}}}{3}+C$
$\Rightarrow -\frac{1}{3}{\left(\frac{1}{x^2-1}\right)}^{\frac{3}{2}}$

$\Rightarrow \frac{(\sqrt{1-x^2}{)}^3}{-3x^2}+C$
$A\left(x\right)=-\frac{1}{3x^3}$ and $m=3$
$\left(A\right(x){)}^m={(-\frac{1}{3x^3})}^3=-\frac{1}{27x^9}$
Case-II $x\le 0$

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