partial fraction derivative question So I have this partial fraction derivative question. I know how to solve it, but for some reason I keep swapping two numbers. Here is the problem:
$$\int\frac{3-4x}{x^2+x}= \frac{A}{x}+\frac{B}{x+1}$$
$$(3-4x)=A(x)+B(x+1)$$
Let $x=-1$
$$(3-4(-1))=A(-1)+B(0)$$
$$-7=A$$
Let $x=0$
$$(3-4(0))=A(0)+B(1)$$
$$3=B$$
Therefore,
$$\int\frac{-7}{x}+\int\frac{3}{x+1}$$
Answer is: $$-7ln|x|+3|x+1|+C$$
shouldn't the answer actually be $$3ln|x|-7|x+1|+C$$
Where am I going wrong?

$$\frac{3-4x}{x^2+x}= \frac{A}{x}+\frac{B}{x+1}$$
$$\frac{A}{x}+\frac{B}{x+1}=\frac {A(x+1)}{x(x+1)}+\frac{Bx}{x(x+1)}$$
then we have to set:$$3-4x=(A+B)x+A$$
 Then $A=3$ and $B=-7$ so our integral will be:
$$\int\frac{3-4x}{x^2+x}dx=\int\frac{3}{x}dx+\int \frac{-7}{x+1}dx $$
$=3\ln\mid x\mid-7\ln\mid(x+1)\mid+c$
