How do you integrate $\frac{2 x}{4 x^{2} + 12 x + 9}$ $(2x) / (4x^2 + 12x + 9)$ using partial fractions?

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How do you integrate $\frac{2 x}{4 x^{2} + 12 x + 9}$ $(2x) / (4x^2 + 12x + 9)$ using partial fractions?

1 Answer

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Answer 1 · 2016-12-28T14:48:49

$= \frac{1}{2} ln | 2 x + 3 | + \frac{3}{2 (2 x + 3)} + C$ $=1/2ln|2x+3|+3/(2(2x+3))+C$ .
You don't use partial fractions, because the denominator is a perfect square.

Explanation:

$\int \frac{2 x}{4 x^{2} + 12 x + 9} d x$ $int (2x)/(4x^2+12x+9)dx$
$= \int \frac{2 x}{{(2 x + 3)}^{2}} d x$ $=int(2x)/(2x+3)^2dx$
So substitute $u = 2 x + 3$ $u=2x+3$ , $d x = \frac{d u}{2}$ $dx=(du)/2$ :
$= \frac{1}{2} \int \frac{u - 3}{u^{2}} d u$ $=1/2int (u-3)/u^2du$
$= \frac{1}{2} \int u^{- 1} - 3 u^{- 2} d x$ $=1/2intu^-1-3u^-2dx$
$= \frac{1}{2} ln | 2 x + 3 | + \frac{3}{2 (2 x + 3)} + C$ $=1/2ln|2x+3|+3/(2(2x+3))+C$

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How do you integrate $\frac{2 x}{4 x^{2} + 12 x + 9}$ $(2x) / (4x^2 + 12x + 9)$ using partial fractions?

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How do you integrate $\frac{2 x}{4 x^{2} + 12 x + 9}$ $(2x) / (4x^2 + 12x + 9)$ using partial fractions?