JEE PYQ: Indefinite Integration Question 30
Question 30 - 2019 (10 Jan Shift 2)
If $\int x^5 e^{-4x^3},dx = \frac{1}{48}e^{-4x^3}f(x) + C$, where $C$ is a constant of integration, then $f(x)$ is equal to:
(1) $-2x^3 - 1$
(2) $-4x^3 - 1$
(3) $-2x^3 + 1$
(4) $4x^3 + 1$
Show Answer
Answer: (2)
Solution
Put $-4x^3 = \theta$, $-12x^2,dx = d\theta$, $x^2,dx = -\frac{d\theta}{12}$. $I = \int \frac{1}{48}\theta e^\theta,d\theta = \frac{1}{48}(\theta e^\theta - e^\theta) + C = \frac{1}{48}e^{-4x^3}(-4x^3-1) + C$. So $f(x) = -4x^3 - 1$.
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