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


 A: Give this one a whirl.
$$\frac{20}{(x-1)(x^2+9)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+9} $$
You should find that $A=2$ and $B=C=-2$. Hence
$$ \int \frac{20}{(x-1)(x^2+9)}\mathrm{d}x = \int \frac{2}{x-1}\mathrm{d}x - \int \frac{2x}{x^2+9}\mathrm{d}x - \int \frac{2}{x^2+9}\mathrm{d}x. $$
A: 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!
A: You made a common mistake with partial fractions. When there is a quadratic denominator, you have to have a linear term in the numerator:
$\dfrac{20}{(x-1)(x^2+9)}=\dfrac{A}{x-1}+\dfrac{\color{blue}{Bx+C}}{x^2+9}.$
Try this and it should work. Since the numerator in the overall fraction is constant, the logarithms will combine nicely and make the antiderivative bounded as $x\to\infty$.
