Please help me for a substitution method to evaluate $\int\frac{dx}{(x+a)^2(x+b)^2}$ [closed]

Question

Please help me evaluate: $$\int\frac{dx}{(x+a)^2(x+b)^2}$$

Answer 1

HINT: Use partial fractions. Move your mouse over the gray area below on how to obtain the partial fraction decomposition.

$$\dfrac1{(x+a)^2(x+b)^2} = \dfrac{A_1}{(x+a)} + \dfrac{A_2}{(x+a)^2} + \dfrac{B_1}{(x+b)} + \dfrac{B_2}{(x+b)^2}$$ This gives us that $$A_1(x+a)(x+b)^2 + A_2(x+b)^2 + B_1(x+b)(x+a)^2 + B_2(x+a)^2 = 1$$ Set $x=-a$ to get $A_2 = \dfrac1{(b-a)^2}$. Set $x=-b$ to get $B_2 = \dfrac1{(a-b)^2}$. Differentiate the above to get $$A_1(x+b)^2 + 2A_1(x+a)(x+b) + 2A_2(x+b) + B_1(x+a)^2 + 2B_1(x+b)(x+a) + 2B_2(x+a) = 0$$ Now set $x=-a$, to get $A_1 = \dfrac{2A_2}{a-b}$. Set $x=-b$, to get $B_1 = \dfrac{2B_2}{b-a}$. Hence, we now have that $$\dfrac1{(x+a)^2(x+b)^2} = \dfrac2{(a-b)^3}\dfrac1{(x+a)} + \dfrac1{(b-a)^2}\dfrac1{(x+a)^2} + \dfrac2{(b-a)^3}\dfrac1{(x+b)} + \dfrac1{(a-b)^2}\dfrac1{(x+b)^2}$$

Answer 2

$$\int\frac{dx}{(x+a)^2(x+b)^2}$$

Since $$1\equiv\left[\frac{(x+a)-(x+b)}{a-b}\right]^2,$$ hence it follows that:

$$\begin{align*} \frac{1}{(x+a)^2(x+b)^2}&=\left[\frac{1}{(x+a)(x+b)}\right]^2\\ &=\left[\frac{\frac{(x+a)-(x+b)}{a-b}}{(x+a)(x+b)}\right]^2\\ &\left[\frac{(x+a)-(x+b)}{(a-b)(x+a)(x+b)}\right]^2\\ &=\frac{(x+a)^2-2(x+a)(x+b)+(x+b)^2}{(a-b)^2(x+a)^2(x+b)^2}\\ &=\frac{1}{(a-b)^2}\left[\frac{1}{(x+b)^2}-\frac{2}{(x+a)(x+b)}+\frac{1}{(x+a)^2}\right]\\ &=\frac{1}{(a-b)^2}\left[\frac{1}{(x+b)^2}-\frac{2}{(a-b)}(\frac{1}{x+b}-\frac{1}{x+a})+\frac{1}{(x+a)^2}\right]\\ &=\frac{1}{(a-b)^2}\left[\frac{(x+a)^2-2(x+a)(x+b)+(x+b)^2}{(x+a)^2(x+b)^2}\right]\\ &=\frac{1}{(a-b)^2}\left[\frac{(x+a)^2}{(x+a)^2(x+b)^2}-\frac{2(x+a)(x+b)}{(x+a)^2(x+b)^2}+\frac{(x+b)^2}{(x+a)^2(x+b)^2}\right]\\ \frac{1}{(x+a)^2(x+b)^2}&=\frac{1}{(a-b)^2}\left[\frac{(x+a)^2}{(x+a)^2(x+b)^2}-\frac{2(x+a)(x+b)}{(x+a)^2(x+b)^2}+\frac{(x+b)^2}{(x+a)^2(x+b)^2}\right]. \end{align*}$$

Now integrate last relation: $$\begin{align*} \int\frac{dx}{(x+a)^2(x+b)^2}&=\frac{1}{(a-b)^2}\int\left[\frac{(x+a)^2}{(x+a)^2(x+b)^2}-\frac{2(x+a)(x+b)}{(x+a)^2(x+b)^2}+\frac{(x+b)^2}{(x+a)^2(x+b)^2}\right]dx\\ &=\frac{1}{(a-b)^2}\int\left[\frac{1}{(x+b)^2}-\frac{2}{a-b}(\frac{1}{x+b}-\frac{1}{x+a})+\frac{1}{(x+b)^2}\right]dx\\ &=\frac{1}{(a-b)^2}\left[\int\frac{dx}{(x+b)^2}-\frac{2}{a-b}\left(\int\frac{dx}{x+b}-\int\frac{dx}{x+a}\right)-\int\frac{dx}{(x+a)^2}\right]\\ &=\frac{1}{(a-b)^2}\left[-\frac{1}{x+b}-\frac{2}{a-b}\ln\left|\frac{x+b}{x+a}\right|-\frac{1}{x+a}\right]\\ &=-\frac{1}{(a-b)^2}\left[\frac{2x+a+b}{(x+a)(x+b)}+\frac{2}{a-b}\ln\left|\frac{x+b}{x+a}\right|\right]. \end{align*}$$

Hence: $$\int\frac{dx}{(x+a)^2(x+b)^2}=-\frac{1}{(a-b)^2}\left[\frac{2x+a+b}{(x+a)(x+b)}+\frac{2}{a-b}\ln\left|\frac{x+b}{x+a}\right|\right]+C.$$

Answer 3

$$\frac{1}{(x + a)^2(x + b)^2} = \left(\frac{1}{\left(x + \frac{a + b}{2}\right)^2 - \left(\frac{a - b}{2}\right)^2}\right)^2$$ Let $y = x + \frac{a + b}{2}$ and $c = \frac{a - b}{2}$, therefore the integral looks like $$\int \frac{1}{(y^2 - c^2)^2}dy$$ Put $y = c\sin(\theta)$, therefore, $$\int \frac{c\cos(\theta)}{c^4\cos^4(\theta)}d\theta = \frac{1}{c^3}\int \sec^3(\theta) d\theta = \frac{1}{2c^3}\left(\sec(\theta)\tan(\theta) + \ln|\sec(\theta) + \tan(\theta)|\right) + C$$ Now, $\sec(\theta) = \frac{1}{\sqrt{1 - \sin^2(\theta)}} = \frac{c}{\sqrt{c^2 - y^2}}$ and $\tan(\theta) = \frac{sin(\theta)}{\sqrt{1 - \sin^2(\theta)}} = \frac{y}{\sqrt{c^2 - y^2}}$, therefore, $$\ln|\sec(\theta) + \tan(\theta)| = \frac{1}{2}\ln\left|\frac{c + y}{c - y}\right| = \frac{1}{2}\ln\left|\frac{x + a}{x + b}\right|$$ Similarly, you can compute the other term as well and get the final expression in terms of $x,a$ and $b$.