Evaluate $\int\frac{8x+20}{5x^2+25x+20}dx$ I tried to solve it and got $\frac{4}{5} \ln(4+5 x+x^2)+C$ as an answer, but my online homework program says it's incorrect.  What did I do wrong?
I pulled out $\frac{4}{5}$ as a constant and saw that the numerator was the derivative of the denominator. So I put the denominator in a natural log.
 A: Substitute $u=5x^2+25x+20$, $du=10x+25$, then$$\int\frac{8x+20}{5x^2+25x+20}dx=\frac {4}{5} \int \frac {1}{u} du = \frac{4}{5} \ln|5(x^2+5x+4)|+C.$$
A: $$\int \frac{8x+20}{5x^2+25x+20} dx=\int \frac{8x+20}{5(x+1)(x+4)} dx$$
$$\frac{8x+20}{5(x+1)(x+4)} =\frac{A}{5(x+1)}+\frac{B}{5(x+4)}=\frac{A(x+4)+B(x+1)}{5(x+1)(x+4)}$$
So:
$$A+B=8 \\ 4A+B=20$$
$$3A=12 \Rightarrow A=4$$
$$B=8-A=4$$
So:
$$\frac{8x+20}{5(x+1)(x+4)}=\frac{4}{5}  \frac{1}{x+1}+\frac{4}{5} \frac{1}{x+4}$$
Therefore:
$$\int \frac{8x+20}{5(x+1)(x+4)} dx=\frac{4}{5} \ln |x+1|+\frac{4}{5} \ln |x+4|+c$$
A: Here is a general method for integrals of this form. 
$$ I = \int \frac{mx±n}{ax^2+bx+d}dx$$  
$$I =   \int \frac{\frac{m}{2a}(2ax+b) + (n - \frac{mb}{2a})}{(x+ \frac{b}{2a})^2 +  (c -\frac{b^2}{4a^2})}dx  = \frac{m}{2a}ln|ax^2 + bx + c| + (n- \frac{mb}{2a}) \int \frac{dx}{ax^2+bx+c}dx$$
depending on the quadratic in the bottom, the second integral will be in the form of either $\frac{1}{a}\arctan(\frac{u}{a})$ + C or $ \frac{1}{2a}ln|\frac{u-a}{u+a}| + C $
