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

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

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Step 1
Given: $$\displaystyle{I}=\int{\frac{{{x}^{{{2}}}+{12}{x}+{12}}}{{{x}^{{{3}}}-{4}{x}}}}{\left.{d}{x}\right.}$$
For evaluating given integral, first we simplify given expression then integrate it
Step 2
So,
$$\displaystyle{I}=\int{\frac{{{x}^{{{2}}}+{12}{x}+{12}}}{{{x}^{{{3}}}-{4}{x}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{\frac{{{x}^{{{3}}}+{12}{x}+{12}}}{{{x}{\left({x}^{{{2}}}-{4}\right)}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{\frac{{{x}^{{{3}}}+{12}{x}+{12}}}{{{x}{\left({x}^{{{2}}}-{2}^{{{2}}}\right)}}}}{\left.{d}{x}\right.}\ \ \ {\left(∵{a}^{{{2}}}-{b}^{{{2}}}={\left({a}+{b}\right)}{\left({a}-{b}\right)}\right)}$$
$$\displaystyle=\int{\frac{{{x}^{{{2}}}+{12}{x}+{12}}}{{{x}{\left({x}+{2}\right)}{\left({x}-{2}\right)}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{\left({\frac{{{5}}}{{{x}-{2}}}}-{\frac{{{3}}}{{{x}}}}-{\frac{{{1}}}{{{x}+{2}}}}\right)}{\left.{d}{x}\right.}\ \ \ {\left(∵\int{\frac{{{\left.{d}{x}\right.}}}{{{x}+{a}}}}={\ln}{\left|{x}+{a}\right|}+{c}\right)}$$
$$\displaystyle={5}{\ln}{\left|{x}-{2}\right|}-{3}{\ln}{\left|{x}\right|}-{\ln}{\left|{x}+{2}\right|}+{c}$$
Hence, given integral can be find as above.
###### Not exactly what you’re looking for?
poleglit3

$$\displaystyle\int{\frac{{{x}^{{{2}}}+{12}{x}+{12}}}{{{x}^{{{3}}}-{4}{x}}}}{\left.{d}{x}\right.}$$
$$\displaystyle{\frac{{{x}^{{{2}}}+{12}{x}+{12}}}{{{x}^{{{3}}}-{4}{x}}}}={\frac{{{A}}}{{{x}}}}+{\frac{{{B}}}{{{x}-{2}}}}+{\frac{{{C}}}{{{x}+{2}}}}$$
$$\displaystyle\Rightarrow{x}^{{{2}}}+{12}{x}+{12}={A}{\left({x}^{{{2}}}-{4}\right)}+{B}\times{\left({x}+{2}\right)}+{C}\times{\left({x}-{2}\right)}$$
put x=0
$$\displaystyle{12}={A}{\left(-{4}\right)}+{B}\times{0}+{C}\times{0}$$
$$\displaystyle\Rightarrow{A}=-{3}$$
put x=2
4+24+12=B2(2+2)
$$\displaystyle\Rightarrow{40}={B}\times{8}$$
$$\Rightarrow$$ B=5
put $$\displaystyle{x}=-{2},{\left(-{2}\right)}^{{{2}}}+{12}{\left(-{2}\right)}+{12}={A}{\left({\left(-{2}\right)}^{{{2}}}-{4}\right)}+{B}{\left(-{2}\right)}{\left({2}-{2}\right)}+{C}{\left(-{2}\right)}{\left(-{2}-{2}\right)}$$
$$\displaystyle\Rightarrow{4}-{24}+{12}={8}{C}$$
$$\displaystyle\Rightarrow-{8}={8}{C}\Rightarrow{C}=-{1}$$
$$\displaystyle{\frac{{{x}^{{{2}}}+{12}{x}+{12}}}{{{x}^{{{3}}}-{4}{x}}}}={\frac{{{5}}}{{{x}-{2}}}}-{\frac{{{1}}}{{{x}+{2}}}}-{\frac{{{3}}}{{{x}}}}$$
$$\displaystyle\int{\frac{{{x}^{{{2}}}+{12}{x}+{12}}}{{{x}^{{{3}}}-{4}{x}}}}{\left.{d}{x}\right.}={5}\int{\frac{{{\left.{d}{x}\right.}}}{{{x}-{2}}}}-\int{\frac{{{\left.{d}{x}\right.}}}{{{x}+{2}}}}-{3}\int{\frac{{{\left.{d}{x}\right.}}}{{{x}}}}$$
$$\displaystyle={5}{\ln{{\left({x}-{2}\right)}}}-{\ln{{\left({x}+{2}\right)}}}-{3}{\ln{{x}}}+{C}$$
$$\displaystyle={\ln{{\frac{{{\left({x}-{2}\right)}^{{{5}}}}}{{{x}^{{{3}}}{\left({x}+{2}\right)}}}}}}+{C}$$

karton

Given:
$$\int \frac{x^{2}+12x+12}{x^{3}-4x}dx \\\int -\frac{3}{x}+\frac{5}{x-2}-\frac{1}{x+2}dx \\-\int \frac{3}{x}dx+\int \frac{5}{x-2}dx-\int \frac{1}{x+2}dx \\-3\ln (|x|)+5\ln(|x-2|)-\ln(|x+2|) \\Answer: \\-3\ln (|x|)+5\ln(|x-2|)-\ln(|x+2|)+C$$