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

JEE Maths question with a full step-by-step solution.

Question

\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+q

equals

3

6

correct

9

12

Solution

Step 1: Look for the

e^{x}(f+f')

pattern. Take

f(x)=\dfrac{1-x^{3}}{1+x^{3}}

. Step 2: Differentiate by the quotient rule:

f'(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'

over the common denominator

(1+x^{3})^{2}

. Writing

f=\dfrac{(1-x^{3})(1+x^{3})}{(1+x^{3})^{2}}=\dfrac{1-x^{6}}{(1+x^{3})^{2}}

f(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

\displaystyle\int e^{x}(f+f')\,dx=e^{x}f

I=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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