Question

Integrate the function w..r. to x : $x^{2}\cdotp e^{3x}$

Answer 1

Solution:

To integrate the given function we must use integration by parts

Formula:
$\int U.V dx=U\int V dx-\int (\frac{dU}{dx}\int V dx)dx$

$\int\:x^{2}.e^{3x}dx=x^{2}\int\:e^{3x} dx-\int (\frac{d\:x^{2}}{dx}\int\:e^{3x} dx)dx\\\\=\frac{x^{2}e^{3x}}{3}-\int\frac{2x\:e^{3x}}{3} dx\\\\ =\frac{x^{2}e^{3x}}{3}-\frac{2}{3}\int\:x\:e^{3x} dx$

Again it is a form of U.V ,so apply again integration by parts

$\int\:xe^{3x}\:dx=x\int e^{3x}\:dx-\int (\frac{d\:x}{dx}\int e^{3x}\:dx)dx\\\\=\frac{xe^{3x}}{3}-\int \frac{e^{3x}}{3}\:dx\\\\ =\frac{xe^{3x}}{3}-\frac{e^{3x}}{9} +C$

Put this value above

$=\frac{x^{2}e^{3x}}{3}-\frac{2}{3} (\frac{xe^{3x}}{3}-\frac{e^{3x}}{9} )+C\\\\= \frac{x^{2}e^{3x}}{3}-\frac{2}{9} xe^{3x}+\frac{2e^{3x}}{27} )+C$

So,

$\int x^{2}.e^{3x} dx=\frac{x^{2}e^{3x}}{3}-\frac{2}{9} xe^{3x}+\frac{2.e^{3x}}{27} +C$