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JEE Main 2019 · 10 Jan · Shift II

JEE Main 2019 Mathematics question on Indefinite Integration

If ∫x5e−4x3dx=148e−4x3f(x)+C\int x^5 e^{-4x^3} dx = \frac{1}{48} e^{-4x^3} f(x) + C, where CC is a constant of integration, then f(x)f(x) is equal to
  1. A.
    −4x3−1-4x^3 - 1
  2. B.
    4x3+14x^3 + 1
  3. C.
    −2x3−1-2x^3 - 1
  4. D.
    −2x3+1-2x^3 + 1
Show answer and solution

Correct option: A

Solution

Substitute t = x³, so dt = 3x² dx. Then x⁵ dx = x³ · x² dx = t · (dt/3).

The integral becomes (1/3)∫t·e^(-4t) dt.

Integrate by parts: (1/3)[t·e^(-4t)/(-4) - ∫e^(-4t)/(-4) dt] = (1/3)[-t·e^(-4t)/4 - e^(-4t)/16] + C = e^(-4t)/48 · (-4t - 1) + C.

Substituting back t = x³: integral = (1/48)e^(-4x³)(-4x³ - 1) + C, so f(x) = -4x³ - 1.

[JEE Main 2019 · 10 Jan · Shift II]

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