Solveeit Logo
Login

Question

Question: The value of the integral \(\int {{e^x}\left[ {\dfrac{{1 + n{x^{n - 1}} - {x^{2n}}}}{{\left( {1 - {x...

The value of the integral ∫ex[1+nxn−1−x2n(1−xn)1−x2n]dx\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) ex[1−xn1+xn]+c{e^x}\left[ {\dfrac{{1 - {x^n}}}{{1 + {x^n}}}} \right] + c
(B) ex[1+xn1−xn]+c{e^x}\left[ {\sqrt {\dfrac{{1 + {x^n}}}{{1 - {x^n}}}} } \right] + c
(C) −ex[1−xn1+xn]+c - {e^x}\left[ {\dfrac{{1 - {x^n}}}{{1 + {x^n}}}} \right] + c
(D) −ex[1+xn1−xn]+c - {e^x}\left[ {\dfrac{{1 + {x^n}}}{{1 - {x^n}}}} \right] + c

Explanation

Solution

Hint : In the given problem, to evaluate the required integration we will use the following formula of integration:
∫ex[f(x)+f′(x)]dx=exf(x)+c\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 ∫ex[1+nxn−1−x2n(1−xn)1−x2n]dx\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 II. Let us simplify this integral by rearranging terms of numerator. So, we can write
I=∫ex[(1−x2n)+nxn−1(1−xn)1−x2n]dxI = \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=∫ex[1−x2n(1−xn)1−x2n+nxn−1(1−xn)1−x2n]dxI = \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−x2n=(1−x2n)(1−x2n)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=∫ex[1−x2n1−xn+nxn−1(1−xn)1−x2n]dx⋯⋯(1)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(x)=1−x2n1−xnf\left( x \right) = \dfrac{{\sqrt {1 - {x^{2n}}} }}{{1 - {x^n}}}. Now we are going to find derivative of f(x)f\left( x \right) by using division rule which is given by ddx[uv]=vdudx−udvdxv2\dfrac{d}{{dx}}\left[ {\dfrac{u}{v}} \right] = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}. Let us take u=1−x2nu = \sqrt {1 - {x^{2n}}} and v=1−xnv = 1 - {x^n}. So, we can write
dudx=ddx[1−x2n]\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left[ {\sqrt {1 - {x^{2n}}} } \right]
⇒dudx=121−x2n(−2nx2n−1)\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{2\sqrt {1 - {x^{2n}}} }}\left( { - 2n{x^{2n - 1}}} \right)
⇒dudx=−nx2n−11−x2n\Rightarrow \dfrac{{du}}{{dx}} = - \dfrac{{n{x^{2n - 1}}}}{{\sqrt {1 - {x^{2n}}} }}
Also we have v=1−xnv = 1 - {x^n}. So, we can write
dvdx=ddx[1−xn]\dfrac{{dv}}{{dx}} = \dfrac{d}{{dx}}\left[ {1 - {x^n}} \right]
⇒dvdx=0−nxn−1\Rightarrow \dfrac{{dv}}{{dx}} = 0 - n{x^{n - 1}}
⇒dvdx=−nxn−1\Rightarrow \dfrac{{dv}}{{dx}} = - n{x^{n - 1}}
Now we can write
f′(x)=ddx[f(x)]f'\left( x \right) = \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]
⇒f′(x)=ddx[1−x2n1−xn]\Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left[ {\dfrac{{\sqrt {1 - {x^{2n}}} }}{{1 - {x^n}}}} \right]
⇒f′(x)=(1−xn)ddx[1−x2n]−(1−x2n)ddx[1−xn](1−xn)2\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}}}
⇒f′(x)=(1−xn)(−nx2n−11−x2n)−(1−x2n)(−nxn−1)(1−xn)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}}}
⇒f′(x)=(1−xn)(−nx2n−1)−(1−x2n)(−nxn−1)(1−xn)21−x2n\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 a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right), we can write 1−x2n=(1)2−(xn)2=(1−xn)(1+xn)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′(x)=(1−xn)(−nx2n−1)−(1−xn)(1+xn)(−nxn−1)(1−xn)21−x2nf'\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 (1−xn)\left( {1 - {x^n}} \right) from numerator and denominator, we get
f′(x)=−nx2n−1+nxn−1+nx2n−1(1−xn)1−x2nf'\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}}} }}
⇒f′(x)=nxn−1(1−xn)1−x2n\Rightarrow f'\left( x \right) = \dfrac{{n{x^{n - 1}}}}{{\left( {1 - {x^n}} \right)\sqrt {1 - {x^{2n}}} }}
Hence, from (1)\left( 1 \right) we can write I=∫ex[f(x)+f′(x)]dx⋯⋯(2)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 ∫ex[f(x)+f′(x)]dx=exf(x)+c\int {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]} dx = {e^x}f\left( x \right) + c. Hence, from (2)\left( 2 \right) we can write
I=exf(x)+c⋯⋯(3)I = {e^x}f\left( x \right) + c \cdots \cdots \left( 3 \right). We have f(x)=1−x2n1−xnf\left( x \right) = \dfrac{{\sqrt {1 - {x^{2n}}} }}{{1 - {x^n}}}. So, from (3)\left( 3 \right) we can write
I=ex[1−x2n1−xn]+cI = {e^x}\left[ {\dfrac{{\sqrt {1 - {x^{2n}}} }}{{1 - {x^n}}}} \right] + c
⇒I=ex[(1−xn)(1+xn)1−xn]+c\Rightarrow I = {e^x}\left[ {\dfrac{{\sqrt {\left( {1 - {x^n}} \right)\left( {1 + {x^n}} \right)} }}{{1 - {x^n}}}} \right] + c
⇒I=ex[1+xn1−xn]+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 ex{e^x}is multiplied with sum of two terms then we can think about the formula ∫ex[f(x)+f′(x)]dx=exf(x)+c\int {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]} dx = {e^x}f\left( x \right) + c where cc is arbitrary (integrating) constant.