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…

If ∫x5e-4x3dx=148e-4x3fx+C, where C is a constant of integration, then fx is equal to
  1. -4x3-1
  2. -2x3+1
  3. -2x3-1
  4. 4x3+1

Solution

We have, I=∫x5e-4x3dx 

Let, -4x3=t⇒-12x2dx=dt

⇒I=∫-t4et-dt12=148∫tetdt

=148tet-∫etdt=t-1et48+C=-4x3-1e-4x348+C, 
where C is the constant of integration.

Hence, fx=-4x3-1

Asked in: JEE Main 2019 (10 Jan Shift 2)

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