Question:
If $\int x^5 e^{-4x^3} dx = \frac{1}{48}e^{-4x^3} f(x) + c$, then $f(x) =$
If $\int x^5 e^{-4x^3} dx = \frac{1}{48}e^{-4x^3} f(x) + c$, then $f(x) =$
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When exponent contains $x^n$, substitution $t=x^n$ simplifies integration quickly.
Updated On: Jun 10, 2026
- $-2x^3 - 1$
- $-4x^3 - 1$
- $-2x^2 + 1$
- $4x^3 + 1$
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The Correct Option is B
Solution and Explanation
Let \( t = x^3 \Rightarrow dt = 3x^2 dx \)
\[
I = \int x^3 \cdot x^2 e^{-4x^3} dx
= \frac{1}{3}\int t e^{-4t} dt
\]
Integration by parts:
\[
I = \frac{1}{3}\left(-\frac{t}{4}e^{-4t} - \frac{1}{16}e^{-4t}\right)
\]
\[
= \frac{1}{48}e^{-4t}(-4t-1)
\]
Substitute \( t=x^3 \):
\[
f(x) = -4x^3 - 1
\]
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