If u and v are the two functions of x, then
$\int_{}^{} (uv) dx = u\left(\int_{}^{} (v)dx\right) - \int_{}^{}\left\{\frac{\text{d}u}{\text{d}x}\int_{}^{}(v) dx\right\}dx $
OR
$\int_{}^{} u.dv= u.v - \int_{}^{} v.du $
Example 1 : Evaluate : $\int_{ }^{ } x.sin(3x) dx$
Solution : Here there are two functions, one is polynomial and 2nd is trigonometric function. So according to LIPTE, u = x and $\frac{\text{d}v}{\text{d}x}$= sin(3x)
u = x
du = dx
$\frac{\text{d}v}{\text{d}x}$= sin(3x)
dv = sin(3x) dx
v= $\int_{ }^{ } sin(3x)dx$
v = $\frac{sin(3x)}{3}$
According to integration by parts,
$\int_{}^{} (uv) dx = u\left(\int_{}^{} (v)dx\right) - \int_{}^{}\left\{\frac{\text{d}u}{\text{d}x}\int_{}^{}(v) dx\right\}dx $
$\int_{}^{}x.sin(3x) dx = x.\int_{}^{}sin(3x)dx - \int_{}^{}\frac{\text{d}}{\text{d}x}(x)\int_{}^{}sin(3x) dx$
= -x $\frac{cos(3x)}{3} - \int_{}^{} \left\{\frac{-cos(3x)}{3}\right\}dx $
= $-x\frac{cos(3x)}{3} + \frac{1}{3}\int_{}^{} cos(3x)dx$
= $-x\frac{cos(3x)}{3} + \frac{1}{3}.\frac{sin(3x)}{3}$
= $-\frac{1}{3}x.cos(3x) + \frac{1}{9} sin(3x) + C $
Example 2: Evaluate $x^{2}.ln(x)$
Solution : Here there are two functions, first is polynomial and 2nd is logarithmic function. So according to LIPTE, u = log(x) and $\frac{\text{d}v}{\text{d}x}= x^{2}$
u = ln(x)
du = $\frac{1}{x} dx $
$\frac{\text{d}v}{\text{d}x}= x^{2}$
dv = $x^{2}$dx
v= $\int_{ }^{ } x^{2}dx$
v = $\frac{x^{3}}{3}$
According to integration by parts,
$\int_{}^{} u.dv= u.v - \int_{}^{} v.du $
$\int_{}^{} ln(x).x^{2} = ln(x)\frac{x^{3}}{3}v - \int_{}^{} \frac{x^{3}}{3}.\frac{1}{x} $
= $\frac{1}{3}.x^{3} ln(x) - \int_{}^{} \frac{x^{2}}{3} $
=$\frac{1}{3}.x^{3} ln(x) - \frac{1}{3} \int_{}^{} x^{2} $
= $\frac{1}{3}.x^{3} ln(x) - \frac{1}{3}. \frac{x^{3}}{3} $
$\int_{}^{} ln(x).x^{2} =\frac{1}{3}.x^{3} ln(x) - \frac{x^{3}}{9} + C $