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Find p+q from an eˣ(f+f') Integral | JEE

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Question
If ∫ex 1−x6−6x2(1+x3)2 dx=ex 1−xp1+xq+c\displaystyle\int e^{x}\,\dfrac{1-x^{6}-6x^{2}}{(1+x^{3})^{2}}\,dx=e^{x}\,\dfrac{1-x^{p}}{1+x^{q}}+c, then p+qp+q equals
A33
B66correct
C99
D1212
Solution
Step 1: Look for the ex(f+f′)e^{x}(f+f') pattern. Take f(x)=1−x31+x3f(x)=\dfrac{1-x^{3}}{1+x^{3}}. Step 2: Differentiate by the quotient rule:
f′(x)=(−3x2)(1+x3)−(1−x3)(3x2)(1+x3)2=−3x2−3x5−3x2+3x5(1+x3)2=−6x2(1+x3)2f'(x)=\dfrac{(-3x^{2})(1+x^{3})-(1-x^{3})(3x^{2})}{(1+x^{3})^{2}}=\dfrac{-3x^{2}-3x^{5}-3x^{2}+3x^{5}}{(1+x^{3})^{2}}=\dfrac{-6x^{2}}{(1+x^{3})^{2}}
 Step 3: Add f+f′f+f' over the common denominator (1+x3)2(1+x^{3})^{2}. Writing f=(1−x3)(1+x3)(1+x3)2=1−x6(1+x3)2f=\dfrac{(1-x^{3})(1+x^{3})}{(1+x^{3})^{2}}=\dfrac{1-x^{6}}{(1+x^{3})^{2}},
f(x)+f′(x)=(1−x6)−6x2(1+x3)2=1−x6−6x2(1+x3)2f(x)+f'(x)=\dfrac{(1-x^{6})-6x^{2}}{(1+x^{3})^{2}}=\dfrac{1-x^{6}-6x^{2}}{(1+x^{3})^{2}}
 which is exactly the bracket in the integrand. Step 4: Apply ∫ex(f+f′) dx=exf\displaystyle\int e^{x}(f+f')\,dx=e^{x}f:
I=ex 1−x31+x3+c  ⇒  p=3, q=3  ⇒  p+q=6I=e^{x}\,\dfrac{1-x^{3}}{1+x^{3}}+c\;\Rightarrow\; p=3,\ q=3\;\Rightarrow\; p+q=6
 Correct answer: (2)
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