Question

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

• A. $\frac{1}{2}[si{n}^{-1}{x}^2+\sqrt{1-x^4}]+c$
• B. $\frac{1}{2}[si{n}^{-1}{x}^2+\sqrt{1-x^2}]+c$
• C. $si{n}^{-1}{x}^2+\sqrt{1-x^4}+c$
• D. $si{n}^{-1}{x}^2+\sqrt{1-x^2}+c$

Answer 1

The correct option is A $\frac{1}{2}[si{n}^{-1}{x}^2+\sqrt{1-x^4}]+c$

$\int x\sqrt{\frac{1-x^2}{1+x^2}}dx=\int \frac{x.(1-x^2)}{\sqrt{1-x^4}}dx$
{Multiplying N' and D' by $(1-x^2{)}^{1/2}$}
$=\int \frac{x}{\sqrt{1-x^4}}dx-\int \frac{x^3}{\sqrt{1-x^4}}dx=\frac{1}{2}\left[si{n}^{-1}\right(x^2)+\sqrt{1-x^4}]+c$ (By putting $x^2=t$ and $\sqrt{1-x^4}=\sqrt{t}$ respectively)