The value of ∫ex1+nxn−1−x2n1−xn1−x2ndx is equal to

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

Solution:

Let $f (x) = \frac{\sqrt{1 - x^{2 n}}}{(1 - x^{n})}$

$\begin{array}{l} \Rightarrow f^{'} (x) = \frac{(\sqrt{1 - x^{n}}) (\frac{1}{2 \sqrt{1 + x^{n}}} . n x^{n - 1}) - (\sqrt{1 + x^{n}}) \frac{1}{2 \sqrt{1 - x^{n}}} . - n x^{n - 1}}{(1 - x^{n})} \\ f' (x) = \frac{(1 - x^{n}) . n x^{n - 1} + (1 + x^{n}) . {nx}^{n - 1}}{2 (1 - x^{n}) \sqrt{1 - x^{2 n}}} \\ f' (x) = \frac{{nx}^{n - 1}}{(1 - x^{n}) \sqrt{1 - x^{2 n}}} \end{array}$

$\begin{array}{l} I = \int \frac{e^{x} (1 + {nx}^{n - 1} - x^{2 n})}{(1 - x^{n}) \sqrt{1 - x^{2 n}}} dx \\ = \int e^{x} \{\frac{\sqrt{1 - x^{2 n}}}{1 - x^{n}} + \frac{{nx}^{n - 1}}{(1 - x^{n}) \sqrt{1 - x^{2 n}}}\} dx \end{array}$

= $\int e^{x} [(f (x) + f' (x)] d x = e^{x} f (x) + c$

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

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

Solution:

Related content