# Thread: [SOLVED] Tech. of Integration

1. ## [SOLVED] Tech. of Integration

xe^(2x)/ (2x+1)^2

2. Originally Posted by swensonm
xe^(2x)/ (2x+1)^2
$\displaystyle \int{\frac{e^{2x}\times x}{(2x+1)^2}}$
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This is a simple question if you remember that
whenever you get an integral with e^x

Try this

$\displaystyle \int{e^{x} (f(x) +f'(x) )} = e^x f(x) + C$

Try proving it its fun
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Here let 2x =t

2dx =dt
dx =dt/2

(2x+1) = t+ 1

$\displaystyle \int{\frac{e^{t}\times t dt }{4(t+1)^2}}$

All you need to do now is See the form , try it before looking down

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Here

$\displaystyle \int{e^{t}~\frac{( t dt) }{4(t+1)^2}}$

$\displaystyle \int{e^{t}~\frac{(t+1-1) dt }{4(t+1)^2}}$

$\displaystyle \int{\frac{e^{t}}{4}(\frac{ t+1}{(t+1)^2}-\frac{ 1 }{(t+1)^2}})dt$

Can you see something
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$\displaystyle \frac{d}{dt} \{\frac{ 1 dt }{(t+1)}\} = \frac{ -1}{(t+1)^2}$

So ultimately
integration has become

$\displaystyle \int{\frac{e^{t}}{4}(f'(t)+f(t))}$

$\displaystyle e^{t} f(t)/4 = \frac{ e^{t} \{ \frac{ 1 }{ (t+1)} \} }{4}$

Put the value of t = 2x

$\displaystyle e^{2x} f(2x)/4 = \{ \frac{ e^{2x} } { 4(2x+1)} \}$

$\displaystyle \int{\frac{e^{2x}\times x}{(2x+1)^2}}$
$\displaystyle \int{e^{x} (f(x) +f'(x) )} = e^x f(x) + C$