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I have a problem with partial decomposition, I got the next function and should integrate the function so it would be easier to split it in two terms. But I have no idea to start this decomposition due to the unknown a.

$\frac{1}{(x+a)(x+1)}$ with $a>0$

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Using Partial Fraction Decomposition

$$\frac1{(x+a)(x+1)}=\frac A{x+a}+\frac B{x+1}$$

Multiplying eithersides by $(x+a)(x+1),$ we get $$1=A(x+1)+B(x+a)=x(A+B)+A+Ba$$

Now compare the constants & the coefficients of $x$


As $x+a-(x+1)=a-1$

by observation $$\frac1{(x+a)(x+1)}=\frac1{a-1}\left(\frac{(x+a)-(x+1)}{(x+a)(x+1)}\right)=\cdots$$

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  • $\begingroup$ I know, but I have problems to find the A and B $\endgroup$ – user2239704 Jan 12 '14 at 16:19
  • $\begingroup$ @user2239704, please find the edited version $\endgroup$ – lab bhattacharjee Jan 12 '14 at 16:23
  • $\begingroup$ With this solution its almost more difficult to get the integral of the solution.. $\endgroup$ – user2239704 Jan 12 '14 at 16:27
  • $\begingroup$ @user2239704, do you mean "Integration"? Then do you know what is $$\int\frac A{x+a}dx$$ $\endgroup$ – lab bhattacharjee Jan 12 '14 at 16:29
  • $\begingroup$ I think it is A*ln(x+a) $\endgroup$ – user2239704 Jan 12 '14 at 16:32
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First Write $\frac{C}{x+a}+\frac{D}{x+1}$ and then solve

$(C+D)x+(C+Da)=1$

$ C+D=0$ and $ C+aD=1$

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  • $\begingroup$ yes but I do not found the C and D for this problem.. I know D = (1-C)/a but I cannot go further $\endgroup$ – user2239704 Jan 12 '14 at 16:24
  • $\begingroup$ $D=-C$ so $C(1-a)=1$ which implies that $C=\frac{1}{1-a}$ and $D=\frac{-1}{1-a}$ So in the end you get $\frac{1}{(x+a)(x+1)}=\frac{1}{1-a}(\frac{1}{x+a}-\frac{1}{x+1})$ $\endgroup$ – b00n heT Jan 12 '14 at 16:26
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Hint $\ $ By Heaviside $\ \dfrac1{(\color{#c00}{x\!-\!a})(\color{#0a0}{x\!-\!b})}=\dfrac A{x-a}+\dfrac B{x-b}\ $ $\Rightarrow$ $\,\ A = \dfrac{1}{a\color{#0a0}{-b}},\ B =\dfrac{1}{b\color{#c00}{-a}} = -A$

Remark $\ $ It deserves to be better known that the Heaviside cover-up method can be generalized to handle nonlinear denominators.

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