Evaluate: $\int \frac{1}{(x+a)(x+b)}$ Evaluate:
$$\int \frac{1}{(x+a)(x+b)}$$
My attempt:
$$\int \frac{1}{(x+a)(x+b)} = \frac{A}{x+a} + \frac{B}{x+b}$$
$$1 = A(x+b) + B(x+a)$$
$$x = -b$$
$$1 = A(-b + b) + B(-b + a)$$
$$1 = B(-b + a)$$
$$x = -a$$
$$1 = A(-a + b) + B(-a + a)$$
$$1 = A(-a + b)$$
At this point I have no idea how to proceed. Can someone help me with this? Please.
 A: You've bagged it. You have
$${1\over (x-a)(x-b)} = {1\over b -a}\left({1\over x + a} - {1\over x  + b }\right).$$
Integrate to obtain
$$\int{dx\over (x-a)(x-b)} = {1\over b -a}(\ln|x + a| - \ln|x + b|) + C$$
Don't become afraid of all of the constants running around.
A: First, it is easy to see there are two cases,(the problem has implied that $x\neq -a,\,-b$)
Case 1, if $a=b$, then the integration is $\int \frac{1}{(x+a)^2}=-\frac{1}{x+a}$.
Case 2, if $a\neq b$, then $\int \frac{1}{(x+a)(x+b)}=\int \frac{1}{b-a}\left(\frac{1}{x+a}-\frac{1}{x+b}\right)=\frac{1}{b-a}\left(\int \frac{1}{x+a}-\int \frac{1}{x+b}\right)=\frac{1}{b-a}\ln \left|\frac{x+a}{x+b}\right|+C$.
A: starting from $1=A(x+b)+B(x+a)=(A+B)x+(Ab+Ba)$, you should get
$$
A+B=0, \quad Ab+Ba=1
$$
and then the solution is
$$
A=1/(b-a),\quad B=1/(a-b)
$$
which means
$$
\frac{1}{(x+a)(x+b)}=\frac{1}{b-a}\left(\frac{1}{x+a}-\frac{1}{x+b}\right)
$$
A: $$\displaystyle {1\over (x+a)(x+b)}={1\over (b-a)}\left[{1\over x+a}-{1\over x+b}\right]$$
