First slide

Methods of integration

Question

$If \int x^{5} e^{4 x^{3}} d x = \frac{1}{48} e^{4 x^{3}} f (x) + C then f (x) =$

Easy

A

$4 x^{3} + 1$
B

$4 x^{3} - 1$
C

$- 4 x^{3} - 1$
D

$2 x^{3} + 1$

Solution

$\begin{array}{l} 4 x^{3} = t \Rightarrow x^{2} d x = \frac{1}{12} d t \\ \begin{array}{ll} \frac{1}{48} \int t e^{t} d t & = \frac{1}{48} e^{t} (t - 1) + C \end{array} \end{array}$
$= \frac{1}{48} e^{4 x^{3}} (4 x^{3} - 1) + C$

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