Question

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

Answer 1

$\int \frac{5x}{(x+1)(x^2-4)}dx=\int \frac{5x}{(x+1)(x+2)(x-2)}dx$
Let $\frac{5x}{(x+1)(x^2-4)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)}......\left(i\right)$
$=\frac{A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2)}{(x+1)(x+2)(x-2)}\Rightarrow 5x=A(x^2-4)+B(x^2+x-2x-2)+C(x^2+x+2x+2)\Rightarrow 5x=x^2(A+B+C)+x(-B+3C)+(-4A-2B+2C)$
On comparing the coefficients of $x^2$,x and constant term on both sides, we get

A+B+C =0 ....(ii)
-B +3C =5....(iii)
and -4A-2B +2C =0 ....(iv)
Multiply by 4 in Eq. (ii) and then adding with Eq. (iv), we get
2(B+3C)=0....(v)
On adding Eqs. (iii)and (v), we get $C=\frac{5}{6}$
On putting the value of C in Eq. (v), we get $B=-\frac{5}{2}$
On putting the values of B and C in Eq. (ii), we get $A=\frac{5}{3}$
$\therefore \int \frac{5x}{(x+1)(x-2)(x+2)}dx=\int \frac{5}{3(x+1)}dx-\int \frac{5}{2(x+2)}dx+\int \frac{5}{6(x-2)}dx=\frac{5}{3}\int \frac{1}{(x+1)}dx-\frac{5}{2}\int \frac{1}{(x+2)}dx+\frac{5}{6}\int \frac{1}{(x-2)}dx=\frac{5}{3}log|x+1|-\frac{5}{2}log|x+2|+\frac{5}{6}log|x-2|+C$