# Integrate the rational functions$\frac{2x-3}{(x^2-1)(2x+3)}$

Toolbox:
• $(i)\;$Form of the rational function$\frac{px+q}{(x-a)(x-b)(x-c)}$
• $\;$Form of the partial function$\frac{A}{(x-a)}+\frac{B}{(x-b}+\frac{C}{(x-c)}$
• $(ii)\;\int\frac{dx}{(x-a)}=log|x-a|+c.$
Given $I=\int\frac{2x-3}{(x^2-1)^(2x+3)}dx=\int\frac{2x-3}{(x-1)(x+1)(2x+3)}.$

$\frac{2x-3}{(x-1)(x+2)(2x+3)}=\frac{A}{(x-1)}+\frac{B}{(x+1)}+\frac{C}{(2x+3)}.$

2x-3=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1).2x-3=A(2x^2+5x+3)+B(2x^2+x-3)+C(x^2-1).$Equating the coefficients of$x^2$, 0=2A+2B+C-----(1) Equating the coefficients of x, 2=5A+B------(2) Equating the constant terms, -3=3A-3B-C-------(3) Multiply equ(2) by 2 and subtract from equ(1) Subtract equ(1) and equ(2) 2A+2B+C=0 10A+2B=4 _________________ 8A+C=-4-------(4) Multiply equ(2) by 3 and add it with equ(3), 3A-3B-C=-3 15A+3B=6 ___________________ 18A-C=3-----(5) Now add equ(4) and equ(5) 18A-C=3 -8A+C=-4 ________________ 10A=-1$A=\frac{-1}{10}$Substituting for A in equ(5) we get, 18(-1/10)-C=3$\frac{-9}{5}-C=3\Rightarrow -C=3+\frac{9}{5}.$C=$\frac{-24}{5}.$substituting for A in equ(2) 5A+B=2. 5(-1/10)+B=2 B=2+$\frac{1}{2}B=\frac{5}{2}$Hence$A=-1/10,B=5/2 and c=-24/5.

Now substituting the values of A,B and C we get,

$\frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{-1}{10(x-1)}+\frac{5}{2(x-1)}-\frac{24}{5(2x+3)}$

Therefore $I=\frac{-1}{10}\int{dx}{(x-1)}+\frac{5}{2}\int\frac{dx}{(x+1)}-\frac{24}{5}\int\frac{dx}{2x+3}.$

Put 2x+3=t.

Therefore 2xdx=dt$\Rightarrow dx=\frac{dt}{2}.$

Substituting this

$I=\frac{-1}{10}\int\frac{dx}{(x-1)}+\frac{5}{2}\int\frac{dx}{(x+1)}-\frac{24}{5}\times\frac{1}{2}\int\frac{dt}{t}.$

On integrating we get,

$\;\;\;=\frac{-1}{10}log|x-1|+\frac{5}{2}log|x+1|-\frac{12}{5}log|t|.$

Substituting for t,

$\int\frac{2x-3}{(x^2-1)(2x+3)}dx=\frac{-1}{10}log|x-1|+\frac{5}{2}log|x+1|-\frac{12}{5}log|2x+3|+c.$
edited Apr 9, 2014