Code:

\begin{align*}
\int\frac{3}{(x+1)(x^2+9)}~\text{d}x& = \int\frac{A}{x+1}+\frac{Bx+C}{x^2+9}~\text{d}x \\
& = Ax^2+9A+Bx^2+Cx+Bx+C \\
& = x^2(1A+1B+0C) = 0 \\
& = x(0A+1B+1C) = 0 \\
& = (9A+0B+1C) = 3 \\
& = \text{rref}\begin{pmatrix}
1~1~0~0 \\
0~1~1~0 \\
9~0~1~3
\end{pmatrix} \\
& = A = \frac{3}{10}~B =-\frac{3}{10}~C=\frac{3}{10} \\
& = \int\frac{\frac{3}{10}}{x+1}+\frac{-\frac{3}{10}x+\frac{3}{10}}{x^2+9}~\text{d}x \\
& = \frac{3}{10}\int\frac{1}{x+1}~-\frac{3}{10}\int\frac{x}{x^2+9}+\frac{3}{10}\int\frac{1}{x^2+9} \\
& = u = x^2+9 ~ \text{d}u ~= 2x \\
& = \frac{3}{10}\int\frac{1}{x+1}~-\frac{3}{20}\int\frac{du}{u}+\frac{1}{30}\int\frac{1}{\frac{x^2}{9}+1} \\
& = w = \frac{x}{3}~\text{d}w ~= \frac{\text{d}x}{3} \\
& = \frac{3}{10}\int\frac{1}{x+1}~-\frac{3}{20}\int\frac{du}{u}+\frac{1}{10}\int\frac{1}{w^2+1} \\
& = \frac{3\ln|x+1|}{10}-\frac{3\ln|x^2+9|}{20}+\frac{\tan^{-1}\left(\frac{x}{3}\right)}{10}+C
\end{align*}

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