Find - -2x-5e2x-2x-33dx

Let I =∫(2x 5)e^2x/(2x 3)^3dx =∫[(2x 3) 2]e^2x/(2x 3)^3dx =∫2x 3/(2x 3)^3.e^2x dx ∫2e^2x/(2x 3)^3dx =∫e^2xdx/(2x 3)^2 amp; amp; ∫2e^2x dx/(2x 3)^3 ...

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Byju's Answer

Standard XII

Mathematics

Special Integrals - 3

Find : ∫2x-5e...

Question

Find : $\int \frac{(2 x - 5) e^{2 x}}{(2 x - 3)^{3}} d x$

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Solution

Let
$I = \int \frac{(2 x - 5) e^{2 x}}{(2 x - 3)^{3}} d x$

$= \int \frac{[(2 x - 3) - 2] e^{2 x}}{(2 x - 3)^{3}} d x$

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

$= \begin{matrix} \int \frac{e^{2 x} d x}{(2 x - 3)^{2}} & - & \int \frac{2 e^{2 x} d x}{(2 x - 3)^{3}} I_{1} & I_{2} \end{matrix}$

Finding, $I_{1} = \int \frac{e^{2 x d x}}{(2 x - 3)^{2}}$

Integrating $I_{1}$ by taking $\frac{1}{(2 x - 3)^{2}}$ as the first function and $e^{2 x}$ as second function.

$= \frac{1}{(2 x - 3)^{2}} . \frac{e^{2 x}}{2} - \int [\frac{d}{d x} (2 x - 3)^{- 2} . \frac{e^{2 x}}{2}] d x$

$= \frac{1}{(2 x - 3)^{2}} . \frac{e^{2 x}}{2} - \int (- 2) (2 x - 3)^{- 2 - 1} .2 . \frac{e^{2 x}}{2} d x$

$I_{1} = \frac{1}{(2 x - 3)^{2}} . \frac{e^{2 x}}{2} + \int \frac{2 e^{2 x}}{(2 x - 3)^{3}} d x + C$

$I_{1} = \frac{e^{2 x}}{2 (2 x - 3)^{2}} + I_{2} + C$

Hence,

$I = I_{1} - I_{2} = \frac{e^{2 x}}{2 (2 x - 3)^{2}} + C$

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Find : ∫(2x−5)e2x(2x−3)3dx

Find : $\int \frac{(2 x - 5) e^{2 x}}{(2 x - 3)^{3}} d x$