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

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

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Try $y=x+a$ or $y=x+b$ - and then follow Marvis. –  Mark Bennet Jul 7 '12 at 19:02
[Low accept rate] discourage some from answering your posts. When a question has been answered to your satisfaction, it's considered good form to click the check mark below the voting arrows for that answer. The person whose answer you accept (you can only accept one) gets a small reputation boost, and you send a signal to others that you're no longer in need of an answer. BTW, it's not too late to go back to your old posts and accept your favorite answers. –  draks ... Jul 11 '12 at 13:52

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 $$\dfrac1{(x+a)^2(x+b)^2} = \dfrac2{(a-b)^3}\dfrac1{(x+a)} + \dfrac1{(b-a)^2}\dfrac1{(x+a)^2} + \dfrac2{(b-a)^3}\dfrac1{(x+b)} + \dfrac1{(a-b)^2}\dfrac1{(x+b)^2}$$

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

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

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$$\frac{1}{(x + a)^2(x + b)^2} = \left(\frac{1}{\left(x + \frac{a + b}{2}\right)^2 - \left(\frac{a - b}{2}\right)^2}\right)^2$$ 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 $b$.
Does to a fair degree accomplish the aim of the title ("a substitution method"). Unless we are prepared to go to the complex numbers, there is a problem if $|y|\lt |c|$, because $|\sec\theta| \ge 1$ for all $\theta$. –  André Nicolas Jul 7 '12 at 21:36
@AndréNicolas True. This problem is immediate once you see the substitution $y = c\sin(\theta)$. This substitution always comes with a caveat that $|y/c| \leq 1$. Also, one can notice that $\int 1/(y \sqrt{(c^2 + y^2)})dy = -(1/c)acosech(y/c) + C$, this gives an alternative to solve the intermediate integral. So, one has at least a couple of options to solve the intermediate integral (just have to be careful with the gotcha's like the one you pointed out). –  TenaliRaman Jul 7 '12 at 22:06