Question

Integrate the rational functions.
$\int \frac{3x+5}{x^3-x^2-x+1}dx.$

Answer 1

$\int \frac{3x+5}{x^3-x^2-x+1}dx=\int \frac{3x+5}{(x^2-1)(x-1)}dx=\int \frac{3x+5}{(x-1)(x+1)(x-1)}dx=\int \frac{3x+5}{(x-1{)}^2(x+1)}dx$
Let $\frac{3x+5}{(x-1{)}^2(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1{)}^2}+\frac{C}{x+1}$
$\Rightarrow \frac{3x+5}{(x-1{)}^2(x+1)}=\frac{A(x-1)(x+1)+B(x+1)+C(x-1{)}^2}{(x-1{)}^2(x+1)}\Rightarrow 3x+5=A(x^2-1)+Bx+B+B+C(x^2+1-2x)\Rightarrow 3x+5=x^2(A+C)+x(B-2C)+(-A+B+C)$
On comparing the coefficients of $x^2$, x and constant term both sides, we get
$A+C=0\Rightarrow A=-C.........\left(i\right)$
$B-2C=3.......\left(ii\right)$
and $-A+B+C=\mathrm{5.......}\left(iii\right)$
On comparing the value of A from Eq. (i) in Eq. (iii), we get
$-(-C)+B+C=5\Rightarrow B+2C=5......\left(iv\right)$
On adding Eq.(ii)and Eq.(iv), we get 2B=8 $\Rightarrow B=4$
On putting the value of B in Eq.(ii), we get $4-2C=3\Rightarrow 1=2C\Rightarrow C=\frac{1}{2}$
Also, $A=-C=-\frac{1}{2}$
$\therefore \int \frac{3x+5}{(x-1{)}^2(x+1)}dx=\int (\frac{-\frac{1}{2}}{x-1}+\frac{4}{(x-1{)}^2}+\frac{\frac{1}{2}}{x+1})dx=-\frac{1}{2}\int \frac{1}{x-1}dx+4\int \frac{1}{(x-1{)}^2}dx+\frac{1}{2}\int \frac{1}{x+1}dx=-\frac{1}{2}log|x-1|+4\frac{(x-1{)}^{-2+1}}{(-2+1)}+\frac{1}{2}log|x+1|+C=\frac{1}{2}log\left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+C[\therefore logb-loga=log(\frac{b}{a}\left)\right]$