Question

The value of $\int e^x\frac{1+n{x}^{n-1}-x^{2n}}{(1-x^n)\sqrt{1-x^{2n}}}dx$ is equal to

• A. $e^x\left(\sqrt{1-x^2}\right)+c$
• B. $e^x\frac{\sqrt{1+x^{2n}}}{1+x^{2n}}+c$
• C. $\frac{e^x\sqrt{1-x^n}}{1-x^{2n}}+c$
• D. $\frac{e^x\sqrt{1-x^{2n}}}{1-x^n}+c$

Answer 1

The correct option is D $\frac{e^x\sqrt{1-x^{2n}}}{1-x^n}+c$

$\int e^x\left(\frac{(1-x^{2n})+n{x}^{n-1}}{(1-x^n)\sqrt{1-x^{2n}}}\right)dx.$
$=\int e^x(\frac{\sqrt{1-x^{2n}}}{1-x^n}+\frac{n{x}^{n-1}}{(1-x^n)\sqrt{1-x^{2n}}})dx$
$\frac{d}{dx}\left(\frac{\sqrt{1-x^{2n}}}{1-x^n}\right)=\frac{-n.x^{2n-1}(1-x^n)+n{x}^{n-1}(1-x^{2n})}{(1-x^n{)}^2\sqrt{(1-x^{2n})}}=\frac{n{x}^{n-1}}{(1-x^n)(\sqrt{1-x^{2n}}}$
It is in the form of $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx$
$\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)+c$
Here $f\left(x\right)=\frac{e^x\sqrt{1-x^{2n}}}{1-x^n}$
$\Rightarrow \int e^x\frac{(1+n{x}^{n-1}-x^{2n})}{(1-x^n)\sqrt{1-x^{2n}}}\cdot dx=\frac{e^x\sqrt{1-x^{2n}}}{1-x^n}+c$