partial fraction derivative question

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?

Answer 1

$(3−4x)=A(x)+B(x+1)$ is wrong.It should be $(3−4x)=A(x+1)+B(x)$

Answer 2

$$\frac{3-4x}{x^2+x}= \frac{A}{x}+\frac{B}{x+1}$$

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

$$\int\frac{3-4x}{x^2+x}dx=\int\frac{3}{x}dx+\int \frac{-7}{x+1}dx $$ $=3\ln\mid x\mid-7\ln\mid(x+1)\mid+c$