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# Integrate the rational functions$\frac{5x}{(x+1)(x^2-4)}$

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 $I=\int\frac{5x}{(x+1)^(x^2-4)}dx=\int\frac{5x}{(x+1)(x+2)(x-2)}dx.$

$\frac{5x}{(x+1)(x+2)(x-2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)}.$

5x=A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2)

$5x=A(x^2-4)+B(x^2-x-2)+C(x^2+3x+2).$

Equating the coefficients of $x^2$,

0=A+B+C-----(1)

Equating the coefficients of x,

5=-B+3C------(2)

Equating the constant terms,

0=-4A-2B+2C-------(3)

A+B+C=0
-B+3C=5
_________________
A+4C=5-------(4)
Multiply equ(2) by 2 and subtract from equ(3),
-2B+6C=10
-4A-2B+2C=0
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4A+4C=10-----(5)
Subtract equ(4) and equ(5)
4A+4C=10
A+4C=5
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3A=5
$A=\frac{5}{3}$
Substituting for A in equ(4) we get,

5/3+4C=5

Therefore 4C=5-5/3=10/3

$\Rightarrow C=\frac{10}{12}=\frac{5}{6}.$

substituting for C in equ(2)

-B+3(5/6)=5

-B=5-5/2

B=$\frac{-5}{2}$

Hence $A=5/3,B=-5/2 and c=5/6. Now substituting the values of A,B and C in I we get, Hence$\int\frac{5x}{(x+1)(x+2)(x-2)}=\frac{5}{3(x+1)}-\frac{5}{2(x+2)}+\frac{5}{6(x-2)}$Therefore$I=\frac{5}{3}\int{dx}{(x+1)}-\frac{5}{2}\int\frac{dx}{(x+2)}+\frac{5}{6}\int\frac{dx}{x-2}.$On integrating we get,$\;\;\;=\frac{5}{3}log|x+1|-\frac{5}{2}log|x+2|+\frac{5}{6}log|x-2|+c.\$