How do I solve $\int \frac{20}{(x-1)(x^2+9)}dx$ I've been trying to solve the following integral:
$\int \frac{20}{(x-1)(x^2+9)}dx$
Sadly I'm kinda new to resolving fractional integrals and I'm not sure which method(s) I should use to solve it.
I've tried using partial fractions but I'm doing something incorrectly or maybe this method isn't the best suited for this case.
I've tried using partial fractions. Here is what I've got so far



We should resolve the integrand as below: $$\frac{20}{(x-1)\left(x^{2}+9\right)} \equiv \frac{A}{x-1}+\frac{B x+C}{x^{2}+9}$$
Then $$20 \equiv A\left(x^{2}+9\right)+(B x+C)(x-1)$$
Putting $x=1$ yields
$$
\begin{aligned}
20=A(10) & \Rightarrow A=2 \\
(B x+C)(x-1) &=20-2\left(x^{2}+9\right) \\
&=2-2 x^{2} \\
&=-2(x+1)(x-1) \\
\therefore \quad B x+C &=-2(x+1) \\
\therefore \frac{20}{(x-1)\left(x^{2}+9\right)} &=\frac{2}{x-1}-\frac{2(x+1)}{x^{2}+9} \\
\int \frac{20}{(x-1)\left(x^{2}+9\right)} d x &=2 \int \frac{d x}{x-1}-\int \frac{2 x d x}{x^{2}+9}-2 \int \frac{d x}{x^{2}+9}
\end{aligned}
$$
Wish it helpful for you to continue!
