Indefinite IntegrationhardFree

Integral of eˣ(1+n·x^(n−1)−x^(2n))/((1−xⁿ)√(1−x^(2n))) | JEE

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Question
∫ex 1+n xn−1−x2n(1−xn)1−x2n dx\displaystyle\int e^{x}\,\dfrac{1+n\,x^{n-1}-x^{2n}}{(1-x^{n})\sqrt{1-x^{2n}}}\,dx equals
Aex1−xn1+xn+Ce^{x}\sqrt{\dfrac{1-x^{n}}{1+x^{n}}}+C
Bex1+xn1−xn+Ce^{x}\sqrt{\dfrac{1+x^{n}}{1-x^{n}}}+Ccorrect
C−ex1−xn1+xn+C-e^{x}\sqrt{\dfrac{1-x^{n}}{1+x^{n}}}+C
D−ex1+xn1−xn+C-e^{x}\sqrt{\dfrac{1+x^{n}}{1-x^{n}}}+C
Solution
Step 1: Look for the ex(f+f′)e^{x}(f+f') pattern. Take f(x)=1+xn1−xnf(x)=\sqrt{\dfrac{1+x^{n}}{1-x^{n}}}. Step 2: Differentiate by logarithms. From ln⁡f=12[ln⁡(1+xn)−ln⁡(1−xn)]\ln f=\dfrac{1}{2}\big[\ln(1+x^{n})-\ln(1-x^{n})\big],
f′f=12[nxn−11+xn+nxn−11−xn]=12⋅nxn−1[(1−xn)+(1+xn)](1+xn)(1−xn)=nxn−11−x2n\dfrac{f'}{f}=\dfrac{1}{2}\left[\dfrac{n x^{n-1}}{1+x^{n}}+\dfrac{n x^{n-1}}{1-x^{n}}\right]=\dfrac{1}{2}\cdot\dfrac{n x^{n-1}\big[(1-x^{n})+(1+x^{n})\big]}{(1+x^{n})(1-x^{n})}=\dfrac{n x^{n-1}}{1-x^{2n}}
 Step 3: Write ff and f′f' over the common denominator (1−xn)1−x2n(1-x^{n})\sqrt{1-x^{2n}}. Since f=1+xn1−x2n=1−x2n(1−xn)1−x2nf=\dfrac{1+x^{n}}{\sqrt{1-x^{2n}}}=\dfrac{1-x^{2n}}{(1-x^{n})\sqrt{1-x^{2n}}} and f′=f⋅nxn−11−x2n=nxn−1(1−xn)1−x2nf'=f\cdot\dfrac{n x^{n-1}}{1-x^{2n}}=\dfrac{n x^{n-1}}{(1-x^{n})\sqrt{1-x^{2n}}},
f+f′=(1−x2n)+nxn−1(1−xn)1−x2n=1+nxn−1−x2n(1−xn)1−x2nf+f'=\dfrac{(1-x^{2n})+n x^{n-1}}{(1-x^{n})\sqrt{1-x^{2n}}}=\dfrac{1+n x^{n-1}-x^{2n}}{(1-x^{n})\sqrt{1-x^{2n}}}
 exactly the integrand's bracket. Step 4: Apply ∫ex(f+f′) dx=exf\displaystyle\int e^{x}(f+f')\,dx=e^{x}f:
I=ex1+xn1−xn+CI=e^{x}\sqrt{\dfrac{1+x^{n}}{1-x^{n}}}+C
 Correct answer: (2)
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