Problems solving partial fraction

Question

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 $a>0$

Answer 1

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 $x$

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

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

Answer 2

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$

Answer 3

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.