## Calculus (3rd Edition)

$$-\frac{1}{5} \ln |x-1|-\frac{1}{x-1}+\frac{1}{10} \ln \left|x^{2}+9\right|-\frac{4}{15} \tan ^{-1}\left(\frac{x}{3}\right)+C$$
Given $$\int \frac{10 d x}{(x-1)^{2}\left(x^{2}+9\right)}$$ Since \begin{aligned} \frac{10}{(x-1)^{2}\left(x^{2}+9\right)}&=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C x+D}{x^{2}+9}\\ &= \frac{A(x-1)\left(x^{2}+9\right)+B\left(x^{2}+9\right)+(C x+D)(x-1)^{2}}{(x-1)^{2}\left(x^{2}+9\right)}\\ 10&=A(x-1)\left(x^{2}+9\right)+B\left(x^{2}+9\right)+(C x+D)(x-1)^{2} \end{aligned} Then$$\text{at } x =1 \ \ \ \ \to B=1$$ By comparing the coefficients, we get \begin{aligned} A+C &=0 \ \ \ \ \text{ coefficient of } x^3\\ 1-A-2 C+D &=0\ \ \text{ coefficient of } x^2\\ 9 A+C-2 D &=0\ \ \text{ coefficient of } x \\ 9-9 A+D &=10\ \ \ \ \text{ coefficient of } x^0\\ \end{aligned} Then $$A=\frac{-1}{5},\ \ \ \ C=\frac{1}{5},\ \ \ \ D=\frac{-1}{5}$$ Hence \begin{aligned} \int \frac{10 d x}{(x-1)^{2}\left(x^{2}+9\right)} &=-\frac{1}{5} \int \frac{d x}{x-1}+\int \frac{d x}{(x-1)^{2}}+\frac{1}{5} \int \frac{x d x}{x^{2}+9}-\frac{4}{5} \int \frac{d x}{x^{2}+9} \\ &=-\frac{1}{5} \ln |x-1|-\frac{1}{x-1}+\frac{1}{10} \ln \left|x^{2}+9\right|-\frac{4}{15} \tan ^{-1}\left(\frac{x}{3}\right)+C \end{aligned}