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Question: Evaluate \(\int {{e^x}(\dfrac{{1 + n{x^{n - 1}} - {x^{2n}}}}{{(1 - {x^n})\sqrt {1 - {x^{2n}}} }})} d...

Evaluate ∫ex(1+nxn−1−x2n(1−xn)1−x2n)dx\int {{e^x}(\dfrac{{1 + n{x^{n - 1}} - {x^{2n}}}}{{(1 - {x^n})\sqrt {1 - {x^{2n}}} }})} dx
A) ex(1−x2n)1+xn+c\dfrac{{{e^x}(1 - {x^{2n}})}}{{1 + {x^n}}} + c
B) ex1−x2n1+xn+c\dfrac{{{e^x}\sqrt {1 - {x^{2n}}} }}{{1 + {x^n}}} + c $$$$
C) ex1−x2n1−xn+c\dfrac{{{e^x}\sqrt {1 - {x^{2n}}} }}{{1 - {x^n}}} + c
D) ex(1−x2n)1−xn+c\dfrac{{{e^x}(1 - {x^{2n}})}}{{1 - {x^n}}} + c

Explanation

Solution

We have the integration which has an exponential function. Firstly we take the function to expect exponential and simplify it we do some rearrangement in the numerator and split the function into two parts. We convert the numerator in the form a2−b2{a^2} - {b^2} of the first part then we apply the formula of a2−b2{a^2} - {b^2}. Also, we break the denominator 1−xn1 - {x^n} in 1−xn×1−xn\sqrt {1 - {x^n}} \times \sqrt {1 - {x^n}} . The factor 1−xn\sqrt {1 - {x^n}} cancels each other. We get a function in the integration with exponential function with text and its derivative has resulted in integration. So we apply this result and solve it.

Complete step-by-step solution:
We have ∫ex1+nxn−1−x2n(1−xn)1−x2ndx\int {{e^x}} \dfrac{{1 + n{x^{n - 1}} - {x^{2n}}}}{{(1 - {x^n})\sqrt {1 - {x^2}^n} }}dx
Let consider
⇒   I=∫ex1+nxn−1−x2n(1−xn)(1−x2n)dx\Rightarrow \,\,\,I = \int {\dfrac{{{e^x}1 + n{x^{n - 1}} - {x^{2n}}}}{{(1 - {x^n})(\sqrt {1 - {x^2}^n} )}}} dx-----------(1)
Here, we have 1+nxn−1−x2n(1−xn)(1−x2n)\dfrac{{1 + n{x^{n - 1}} - {x^2}^n}}{{(1 - {x^n})(\sqrt {1 - {x^2}^n} )}}
We will simplify this first
⇒  1+nxn−1−x2n(1−xn)(1−x2n)\Rightarrow \,\,\dfrac{{1 + n{x^{n - 1}} - {x^2}^n}}{{(1 - {x^n})(\sqrt {1 - {x^2}^n} )}}
Rearranging the numerator,we get
⇒  1−x2n+nxn−1(1−xn)(1−x2n)\Rightarrow \,\,\dfrac{{1 - {x^2}^n + n{x^{n - 1}}}}{{(1 - {x^n})(\sqrt {1 - {x^2}^n} )}}
Separating the factors we get
⇒  1−x2n(1−xn)(1−x2n)+nxn−1(1−xn)(1−x2n)\Rightarrow \,\,\dfrac{{1 - {x^2}^n}}{{(1 - {x^n})(\sqrt {1 - {x^2}^n} )}} + \dfrac{{n{x^{n - 1}}}}{{(1 - {x^n})(\sqrt {1 - {x^2}^n} )}}
Numerator of first fraction can be written as
⇒  1−x2n⋅1−x2n(1−xn)(1−x2n)+nxn−1(1−xn)(1−x2n)\Rightarrow \,\,\dfrac{{\sqrt {1 - {x^2}^n} \cdot \sqrt {1 - {x^2}^n} }}{{(1 - {x^n})(\sqrt {1 - {x^2}^n} )}} + \dfrac{{n{x^{n - 1}}}}{{(1 - {x^n})(\sqrt {1 - {x^2}^n} )}}
On simplification, we get
⇒  1−x2n1−xn+nxn−1(1−xn)1−x2n\Rightarrow \,\,\dfrac{{\sqrt {1 - {x^2}^n} }}{{1 - {x^n}}} + \dfrac{{n{x^{n - 1}}}}{{(1 - {x^n})\sqrt {1 - {x^2}^n} }}
⇒   (1)2−(xn)21−xn+nxn−1(1−xn)1−x2n\Rightarrow \,\,\,\dfrac{{\sqrt {{{(1)}^2} - {{({x^n})}^2}} }}{{1 - {x^n}}} + \dfrac{{n{x^{n - 1}}}}{{(1 - {x^n})\sqrt {1 - {x^2}^n} }}
Now we have (1)2−(xn)2{(1)^2} - {({x^n})^2}we can apply formula a2−b2=(a+b)(a−b){a^2} - {b^2} = (a + b)(a - b) and also 1−xn1 - {x^n} can be written as 1−x2×1−x2\sqrt {1 - {x^2}} \times \sqrt {1 - {x^2}} , then
⇒   1+xn×1−xn1−xn×1−xn+nxn−1(1−xn)1−x2n\Rightarrow \,\,\,\dfrac{{\sqrt {1 + {x^n}} \times \sqrt {1 - {x^n}} }}{{\sqrt {1 - {x^n}} \times \sqrt {1 - {x^n}} }} + \dfrac{{n{x^{n - 1}}}}{{(1 - {x^n})\sqrt {1 - {x^2}^n} }}
Cancel 1−xn\sqrt {1 - {x^n}} in both numerator and denominator from the first fraction
⇒  1+xn1−xn+nxn−1(1−xn)1−x2n\Rightarrow \,\,\dfrac{{\sqrt {1 + {x^n}} }}{{\sqrt {1 - {x^n}} }} + \dfrac{{n{x^{n - 1}}}}{{(1 - {x^n})\sqrt {1 - {x^2}^n} }}
or
⇒   1+xn1−xn+nxn−1(1−xn)1−x2n\Rightarrow \,\,\,\sqrt {\dfrac{{1 + {x^n}}}{{1 - {x^n}}}} + \dfrac{{n{x^{n - 1}}}}{{(1 - {x^n})\sqrt {1 - {x^2}^n} }}
Then equation (1) becomes
⇒    I=∫ex(1+xn1−xn+nxn−1(1−xn)1−x2n)dx\Rightarrow \,\,\,\,I = \int {{e^x}\left( {\sqrt {\dfrac{{1 + {x^n}}}{{1 - {x^n}}} + } \dfrac{{n{x^{n - 1}}}}{{(1 - {x^n})\sqrt {1 - {x^2}^n} }}} \right)} dx
Now 1+xn1−xn\sqrt {\dfrac{{1 + {x^n}}}{{1 - {x^n}}}} is a function and nxn−1(1−xn)1−x2n\dfrac{{n{x^{n - 1}}}}{{(1 - {x^n})\sqrt {1 - {x^2}^n} }} is derivative
And we know that ∫ex(f(x)+f1(x)) dx=exf(x)+c\int {{e^x}\left( {f(x) + {f^1}(x)} \right)} \,dx = {e^x}f(x) + c
Therefore
⇒   I=∫ex(1+xn1−xn+nxn−1(1−x2)1−x2n)dx\Rightarrow \,\,\,I = \int {{e^x}} \left( {\sqrt {\dfrac{{1 + {x^n}}}{{1 - {x^n}}} + } \dfrac{{n{x^{n - 1}}}}{{(1 - {x^2})\sqrt {1 - {x^2}^n} }}} \right)dx
⇒   I=ex1+xn1−xn+c\Rightarrow \,\,\,I = {e^x}\sqrt {\dfrac{{1 + {x^n}}}{{1 - {x^n}}}} + c
⇒    I=ex1+xn×1−xn1−xn×1−xn+c\Rightarrow \,\,\,\,I = {e^x}\dfrac{{\sqrt {1 + {x^n}} \times \sqrt {1 - {x^n}} }}{{\sqrt {1 - {x^n}} \times \sqrt {1 - {x^n}} }} + c
⇒    I=ex(1+xn)(1−xn)(1−xn)2+c\Rightarrow \,\,\,\,I = {e^x}\dfrac{{\sqrt {\left( {1 + {x^n}} \right)\left( {1 - {x^n}} \right)} }}{{{{\left( {\sqrt {1 - {x^n}} } \right)}^2}}} + c
Apply the formula (a2−b2)=(a+b)(a−b)\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right) in numerator
⇒    I=ex12−x2n1−xn+c\Rightarrow \,\,\,\,I = {e^x}\dfrac{{\sqrt {{1^2} - {x^{2n}}} }}{{1 - {x^n}}} + c

Note: Integration is a way of adding slices to find the whole integration can be used to find the area, volume, and central points. It is used to find many useful quantities.
i) Unit of a function: The unit of a function is a fundament of concepts in calculus and analysis concerning the behaviour of a function near a particular input.
ii) Differentiation: The derivative of a function of a real variable measures the sensitivity to the change of a function with respect to change in argument.