Possible evaluation of an indefinite integral $\int{1\over (x+a)^2\cdot(x+b)^2}$

Question

Taking the integral:

$$\int{1\over (x+a)^2\cdot(x+b)^2}$$ I tried to rewrite it in such a way: $$\int{1\over a-b}\cdot{1\over 2x+a+b}\cdot{(x+a)^2-(x+b)^2\over (x+a)^2\cdot(x+b)^2}$$ I have no idea what to do next. Could you help me to evaluate the indefinite integral?

Answer 1

Notice, $$\int \frac{1}{(x+a)^2(x+b)^2}\ dx$$$$=\int \left(\frac{1}{(x+a)(x+b)}\right)^2\ dx$$ $$=\frac1{(b-a)^2}\int \left(\frac{1}{x+a}-\frac{1}{x+b}\right)^2\ dx$$ $$=\frac 1{(b-a)^2}\int \left(\frac{1}{(x+a)^2}+\frac{1}{(x+b)^2}-\frac{2}{(x+a)(x+b)}\right)\ dx$$ $$=\frac 1{(b-a)^2}\int \left(\frac{1}{(x+a)^2}+\frac{1}{(x+b)^2}-\frac{2}{b-a}\left(\frac{1}{x+a}-\frac{1}{x+b}\right)\right)\ dx$$ $$=\frac 1{(b-a)^2} \left(-\frac{1}{(x+a)}-\frac{1}{(x+b)}-\frac{2}{b-a}\ln\left|\frac{x+a}{x+b}\right|\right)+C$$

$$=-\frac {2x+a+b}{(b-a)^2(x+a)(x+b)}-\frac{2}{(b-a)^3}\ln\left|\frac{x+a}{x+b}\right|+C$$

Answer 2

HINT:

Use Partial Fraction Decomposition,

$$\dfrac1{(x-a)^m(x-b)^n}$$

$$=\dfrac{a_1}{x-a}+\dfrac{a_2}{(x-a)^2}+\cdots+\dfrac{a_m}{(x-a)^m} +\dfrac{b_1}{x-b}+\dfrac{b_2}{(x-b)^2}+\cdots+\dfrac{b_n}{(x-b)^n}$$

where $a_i,b_j;1\le i\le m,1\le j\le n$ are arbitrary constants

Answer 3

Here's an alternative to partial fractions method, $$I(a,b)=\int \frac{dx}{(x-a)(x-b)}=\frac{1}{b-a}\int \left(\frac{1}{x-b}-\frac{1}{x-a}\right)dx=\frac{1}{b-a}(\ln|x-b|-\ln|x-a|)+C$$

Now differentiate w.r.t a and b,$$\frac{d}{da}\frac{d I}{db}=\int \frac{dx}{(x-a)^2(x-b)^2}$$