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In this section, ask-math explains you the important technique of integration called
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 $**

The integral of the products of the two functions

= ( 1st function)X (integral of the 2nd function) - integral of (derivative of 1st function)x ((integral of the 2nd function)

$\frac{\text{d}}{\text{d}x}\left\{f(x).g(x)\right\}= f(x).\frac{\text{d}}{\text{d}x}[g(x)] + g(x).\frac{\text{d}}{\text{d}x}[f(x)]$

$\int_{}^{} \left\{ f(x).\frac{\text{d}}{\text{d}x}[g(x)] + g(x).\frac{\text{d}}{\text{d}x}[f(x)]\right\}dx = f(x).g(x)$

$\Rightarrow\int_{}^{} \left\{ f(x).\frac{\text{d}}{\text{d}x}[g(x)] \right\}+\int_{}^{} \left\{ g(x).\frac{\text{d}}{\text{d}x}[f(x)]\right\}dx = f(x).g(x)$

$\Rightarrow\int_{}^{} \left\{ f(x).\frac{\text{d}}{\text{d}x}[g(x)] \right\}dx = f(x).g(x) - \int_{}^{} \left\{ g(x).\frac{\text{d}}{\text{d}x}[f(x)]\right\}dx$

Let f(x) = u and $\frac{\text{d}}{\text{d}x}[g(x)] dx$ = v So that g(x) = $\int_{}^{}(v) dx$

$ \therefore \int_{}^{} (uv) dx = u\left(\int_{}^{} (v)dx\right) - \int_{}^{}\left\{\frac{\text{d}u}{\text{d}x}\int_{}^{}(v) dx\right\}dx $

Like PEMDAS or BODMAS, in Integration by parts we have to follow rule

I - stands for inverse functions

P - stands for polynomial functions

E - stands for exponential functions

T - stands for trigonometric functions

**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 $

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