Question

The value of the integral $\int {{e^x}\left[ {\dfrac{{1 + n{x^{n - 1}} - {x^{2n}}}}{{\left( {1 - {x^n}} \right)\sqrt {1 - {x^{2n}}} }}} \right]dx} $ is equal to
(A) ${e^x}\left[ {\dfrac{{1 - {x^n}}}{{1 + {x^n}}}} \right] + c$
(B) ${e^x}\left[ {\sqrt {\dfrac{{1 + {x^n}}}{{1 - {x^n}}}} } \right] + c$
(C) $ - {e^x}\left[ {\dfrac{{1 - {x^n}}}{{1 + {x^n}}}} \right] + c$
(D) $ - {e^x}\left[ {\dfrac{{1 + {x^n}}}{{1 - {x^n}}}} \right] + c$

Answer 1

Hint: In the given problem, to evaluate the required integration we will use the following formula of integration:
$\int {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]} dx = {e^x}f\left( x \right) + c$.
Before using this formula, first we will simplify the given integral. We must know the division rule for differentiation.

Complete step-by-step answer:
In this problem, we need to evaluate the integral $\int {{e^x}\left[ {\dfrac{{1 + n{x^{n - 1}} - {x^{2n}}}}{{\left( {1 - {x^n}} \right)\sqrt {1 - {x^{2n}}} }}} \right]} dx$. Let us say this integral is denoted by $I$. Let us simplify this integral by rearranging terms of numerator. So, we can write
$I = \int {{e^x}\left[ {\dfrac{{\left( {1 - {x^{2n}}} \right) + n{x^{n - 1}}}}{{\left( {1 - {x^n}} \right)\sqrt {1 - {x^{2n}}} }}} \right]dx} $
Let us separate the numerator with the denominator in above integral. So, we get
$I = \int {{e^x}\left[ {\dfrac{{1 - {x^{2n}}}}{{\left( {1 - {x^n}} \right)\sqrt {1 - {x^{2n}}} }} + \dfrac{{n{x^{n - 1}}}}{{\left( {1 - {x^n}} \right)\sqrt {1 - {x^{2n}}} }}} \right]} dx$
Let us simplify the above integral by writing $1 - {x^{2n}} = \left( {\sqrt {1 - {x^{2n}}} } \right)\left( {\sqrt {1 - {x^{2n}}} } \right)$ in first term of the RHS. So, we get $I = \int {{e^x}\left[ {\dfrac{{\sqrt {1 - {x^{2n}}} }}{{1 - {x^n}}} + \dfrac{{n{x^{n - 1}}}}{{\left( {1 - {x^n}} \right)\sqrt {1 - {x^{2n}}} }}} \right]} dx \cdots \cdots \left( 1 \right)$
Let us say $f\left( x \right) = \dfrac{{\sqrt {1 - {x^{2n}}} }}{{1 - {x^n}}}$. Now we are going to find derivative of $f\left( x \right)$ by using division rule which is given by $\dfrac{d}{{dx}}\left[ {\dfrac{u}{v}} \right] = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$. Let us take $u = \sqrt {1 - {x^{2n}}} $ and $v = 1 - {x^n}$. So, we can write
$\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left[ {\sqrt {1 - {x^{2n}}} } \right]$
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{2\sqrt {1 - {x^{2n}}} }}\left( { - 2n{x^{2n - 1}}} \right)$
$ \Rightarrow \dfrac{{du}}{{dx}} = - \dfrac{{n{x^{2n - 1}}}}{{\sqrt {1 - {x^{2n}}} }}$
Also we have $v = 1 - {x^n}$. So, we can write
$\dfrac{{dv}}{{dx}} = \dfrac{d}{{dx}}\left[ {1 - {x^n}} \right]$
$ \Rightarrow \dfrac{{dv}}{{dx}} = 0 - n{x^{n - 1}}$
$ \Rightarrow \dfrac{{dv}}{{dx}} = - n{x^{n - 1}}$
Now we can write
$f'\left( x \right) = \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]$
$ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left[ {\dfrac{{\sqrt {1 - {x^{2n}}} }}{{1 - {x^n}}}} \right]$
$ \Rightarrow f'\left( x \right) = \dfrac{{\left( {1 - {x^n}} \right)\dfrac{d}{{dx}}\left[ {\sqrt {1 - {x^{2n}}} } \right] - \left( {\sqrt {1 - {x^{2n}}} } \right)\dfrac{d}{{dx}}\left[ {1 - {x^n}} \right]}}{{{{\left( {1 - {x^n}} \right)}^2}}}$
$ \Rightarrow f'\left( x \right) = \dfrac{{\left( {1 - {x^n}} \right)\left( { - \dfrac{{n{x^{2n - 1}}}}{{\sqrt {1 - {x^{2n}}} }}} \right) - \left( {\sqrt {1 - {x^{2n}}} } \right)\left( { - n{x^{n - 1}}} \right)}}{{{{(1 - {x^n})}^2}}}$
$ \Rightarrow f'\left( x \right) = \dfrac{{\left( {1 - {x^n}} \right)\left( { - n{x^{2n - 1}}} \right) - \left( {1 - {x^{2n}}} \right)\left( { - n{x^{n - 1}}} \right)}}{{{{\left( {1 - {x^n}} \right)}^2}\sqrt {1 - {x^{2n}}} }}$
By using formula ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, we can write $1 - {x^{2n}} = {\left( 1 \right)^2} - {\left( {{x^n}} \right)^2} = \left( {1 - {x^n}} \right)\left( {1 + {x^n}} \right)$. Use this in the numerator, we get
$f'\left( x \right) = \dfrac{{\left( {1 - {x^n}} \right)\left( { - n{x^{2n - 1}}} \right) - \left( {1 - {x^n}} \right)\left( {1 + {x^n}} \right)\left( { - n{x^{n - 1}}} \right)}}{{{{\left( {1 - {x^n}} \right)}^2}\sqrt {1 - {x^{2n}}} }}$
By cancelling the factor $\left( {1 - {x^n}} \right)$ from numerator and denominator, we get
$f'\left( x \right) = \dfrac{{ - n{x^{2n - 1}} + n{x^{n - 1}} + n{x^{2n - 1}}}}{{\left( {1 - {x^n}} \right)\sqrt {1 - {x^{2n}}} }}$
$ \Rightarrow f'\left( x \right) = \dfrac{{n{x^{n - 1}}}}{{\left( {1 - {x^n}} \right)\sqrt {1 - {x^{2n}}} }}$
Hence, from $\left( 1 \right)$ we can write $I = \int {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]} dx \cdots \cdots \left( 2 \right)$. There is a formula of integration which is given by $\int {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]} dx = {e^x}f\left( x \right) + c$. Hence, from $\left( 2 \right)$ we can write
$I = {e^x}f\left( x \right) + c \cdots \cdots \left( 3 \right)$. We have $f\left( x \right) = \dfrac{{\sqrt {1 - {x^{2n}}} }}{{1 - {x^n}}}$. So, from $\left( 3 \right)$ we can write
$I = {e^x}\left[ {\dfrac{{\sqrt {1 - {x^{2n}}} }}{{1 - {x^n}}}} \right] + c$
$ \Rightarrow I = {e^x}\left[ {\dfrac{{\sqrt {\left( {1 - {x^n}} \right)\left( {1 + {x^n}} \right)} }}{{1 - {x^n}}}} \right] + c$
$ \Rightarrow I = {e^x}\left[ {\sqrt {\dfrac{{1 + {x^n}}}{{1 - {x^n}}}} } \right] + c$

So, the correct answer is “Option B”.

Note: In this type of problems, when ${e^x}$is multiplied with sum of two terms then we can think about the formula $\int {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]} dx = {e^x}f\left( x \right) + c$ where $c$ is arbitrary (integrating) constant.