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Evaluate \(\int \frac{x}{x^{2}+5 x+6} d x\)

\(x^{2}+5 x+6=(x+2)(x+3)\)

\(f(x)=\frac{x}{(x+2)(x+3)}=\frac{A}{x+2}+\frac{B}{x+3}\)

\(A=f(x)\left.(x+2)\right|_{x=-2}=(x+2) \cdot\left.\frac{x}{(x+2)(x+3)}\right|_{x=-2}\)

\(=\frac{-2}{-2+3}=\frac{-2}{1}=-2\)

\(B=(x+3) \cdot f\left.(x)\right|_{x=-3}=(x+3) \cdot\left.\frac{x}{(x+2)(x+3)}\right|_{x=-3}\)

\(=\frac{x}{x+2} |_{x=-3}=\frac{-3}{-3+2}=\frac{-3}{-1}=3\)

\(\int f(x) dx=\int \frac{-2}{x+2} d x+\int \frac{3}{x+3} d x\)

\(\int \frac{x}{x^{2}+5 x+6} d x=-2 \int \frac{1}{x+2} d x+3 \int \frac{1}{x+3} d x\)

\(\int \frac{x}{x^{2}+5 x+6} d x=-2 \ln |x+2|+3 \ln |x+3|+c\)

\(\ln a-\ln b=\ln \frac{a}{b}\)

\(a \ln |b|=\ln | {b^{a}}|\)

\(\int \frac{x}{x^{2}+5 x+6} d x=\ln \left|\frac{(x+3)^{3}}{(x+2)^{2}}\right|+c\)

Evaluate \(\int \frac{6}{x^{2}+x} d x\)

\(=6 \int \frac{1}{x^{2}+x} d x=6 \int \frac{1}{x(x+1)} d x\)

\(6 \int \frac{1}{x(x+1)} d x=6\left[\int \frac{A}{x}dx+\int \frac{B}{x+1} d x\right]\)

\(A=x \cdot f\left.(x)\right|_{x=0}=x \cdot \frac{1}{x(x+1)}\)

\(=\left.\frac{1}{x+1}\right|_{x=0}=1\)

\(B=(x+1) \cdot\left.\frac{1}{x(x+1)}\right|_{x=-1}\)

\(=\left.\frac{1}{x}\right|_{x=-1}=-1\)

\(6 \int \frac{1}{x(x+1)} d x=6\left[\int \frac{1}{x} d x+\int \frac{-1}{x+1} d x\right]\)

\(=6[\ln |x|+-\ln |x+1|+c]\)

\(=6 \ln |x|-6 \ln |x+1|+c\)

\(a \ln |b|=\ln | {b^{a}}|\)

\(\ln a-\ln b=\ln \frac{a}{b}\)

\(\int \frac{6}{x^{2}+x}=\ln \left|\frac{x^{6}}{ (x+1)^{6}}\right|+c\)

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