piarepm

2022-01-07

Use partial fractions to find the indefinite integral.
$\int \frac{{x}^{2}+12x+12}{{x}^{3}-4x}dx$

Bob Huerta

Expert

Step 1
Given: $I=\int \frac{{x}^{2}+12x+12}{{x}^{3}-4x}dx$
For evaluating given integral, first we simplify given expression then integrate it
Step 2
So,
$I=\int \frac{{x}^{2}+12x+12}{{x}^{3}-4x}dx$
$=\int \frac{{x}^{3}+12x+12}{x\left({x}^{2}-4\right)}dx$

$=\int \frac{{x}^{2}+12x+12}{x\left(x+2\right)\left(x-2\right)}dx$

$=5\mathrm{ln}|x-2|-3\mathrm{ln}|x|-\mathrm{ln}|x+2|+c$
Hence, given integral can be find as above.

poleglit3

Expert

$\int \frac{{x}^{2}+12x+12}{{x}^{3}-4x}dx$
$\frac{{x}^{2}+12x+12}{{x}^{3}-4x}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+2}$
$⇒{x}^{2}+12x+12=A\left({x}^{2}-4\right)+B×\left(x+2\right)+C×\left(x-2\right)$
put x=0
$12=A\left(-4\right)+B×0+C×0$
$⇒A=-3$
put x=2
4+24+12=B2(2+2)
$⇒40=B×8$
$⇒$ B=5
put $x=-2,{\left(-2\right)}^{2}+12\left(-2\right)+12=A\left({\left(-2\right)}^{2}-4\right)+B\left(-2\right)\left(2-2\right)+C\left(-2\right)\left(-2-2\right)$
$⇒4-24+12=8C$
$⇒-8=8C⇒C=-1$
$\frac{{x}^{2}+12x+12}{{x}^{3}-4x}=\frac{5}{x-2}-\frac{1}{x+2}-\frac{3}{x}$
$\int \frac{{x}^{2}+12x+12}{{x}^{3}-4x}dx=5\int \frac{dx}{x-2}-\int \frac{dx}{x+2}-3\int \frac{dx}{x}$
$=5\mathrm{ln}\left(x-2\right)-\mathrm{ln}\left(x+2\right)-3\mathrm{ln}x+C$
$=\mathrm{ln}\frac{{\left(x-2\right)}^{5}}{{x}^{3}\left(x+2\right)}+C$

karton

Expert

Given:
$\int \frac{{x}^{2}+12x+12}{{x}^{3}-4x}dx\phantom{\rule{0ex}{0ex}}\int -\frac{3}{x}+\frac{5}{x-2}-\frac{1}{x+2}dx\phantom{\rule{0ex}{0ex}}-\int \frac{3}{x}dx+\int \frac{5}{x-2}dx-\int \frac{1}{x+2}dx\phantom{\rule{0ex}{0ex}}-3\mathrm{ln}\left(|x|\right)+5\mathrm{ln}\left(|x-2|\right)-\mathrm{ln}\left(|x+2|\right)\phantom{\rule{0ex}{0ex}}Answer:\phantom{\rule{0ex}{0ex}}-3\mathrm{ln}\left(|x|\right)+5\mathrm{ln}\left(|x-2|\right)-\mathrm{ln}\left(|x+2|\right)+C$

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