Mathematics · 2019
JEE Main · 10 January 2019, Shift 2 · Q74
If ∫ x^5 e^-4 x^3 d x=1/48 e^-4 x^3 f(x)+C, where C is a constant of integration, then f(x) is equal to:
If $\displaystyle \int x^5 e^{-4 x^3} d x=\frac{1}{48} e^{-4 x^3} f(x)+C$,
where $\displaystyle C$ is a constant of integration, then $\displaystyle f(x)$ is equal to:
Official answer
(2)
$\displaystyle -4 x^3-1$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.