Question

Evaluate $\int e^x\left(\frac{1+n{x}^{n-1}-x^{2n}}{(1-x^n)\sqrt{1-x^{2n}}}\right)dx$

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

Answer 1

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

$\int e^x\left(\frac{1+n{x}^{n-1}-x^{2n}}{(1-x^n)\sqrt{1-x^{2n}}}\right)dx$

$\int e^x(\frac{1-x^{2n}}{(1-x^n)\left(\sqrt{1-x^{2n}}\right)}+\frac{n{x}^{n-1}}{(1-x^n)\left(\sqrt{1-x^{2n}}\right)}dx$

$\int e^x(\frac{\sqrt{1-x^{2n}}}{1-x^n}+\frac{n{x}^{n-1}}{(1-x^n)\sqrt{1-x^{2n}}})dx$

$\frac{d}{dx}\left(\frac{\sqrt{1-x^{2n}}}{1-x^n}\right)=\frac{n{x}^{n-1}}{(1-x^n)\sqrt{1-x^{2n}}}$

So, it is in the form of

$\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)+c$

Hence, $\int e^x\left(\frac{1+n{x}^{n-1}-x^{2n}}{(1-x^n)\sqrt{1-x^{2n}}}\right)dx$ $=e^x\left(\frac{\sqrt{1-x^{2n}}}{1-x^n}\right)+c.$