Question

Evaluate:$\int \frac{x^2+1}{{(x-1)}^2(x+3)}dx$

Answer 1

$\int \frac{x^2+1}{{(x-1)}^2(x+3)}dx$

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

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

Put $x=1\Rightarrow 2=0+0+4C\Rightarrow C=\frac{2}{4}=\frac{1}{2}$

Put $x=-3\Rightarrow 9+1=B{(-3-1)}^2\Rightarrow 16B=10$ or $B=\frac{10}{16}=\frac{5}{8}$

Put $x=0\Rightarrow 1=-3A+B+3C\Rightarrow 1=-3A+\frac{5}{8}+\frac{3}{2}$

$\Rightarrow -3A+\frac{5+12}{8}=1$

$\Rightarrow -3A=1-\frac{17}{8}=\frac{8-17}{8}=\frac{-9}{8}$

$\Rightarrow A=\frac{3}{8}$

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

$=\frac{3}{8}\int \frac{dx}{x-1}+\frac{5}{8}\int \frac{dx}{x+3}+\frac{1}{2}\int \frac{dx}{{(x-1)}^2}$

$=\frac{3}{8}\log |x-1|+\frac{5}{8}\log |x+3|+\frac{1}{2}\frac{{(x-1)}^{-2+1}}{-2+1}+c$

$=\frac{3}{8}\log |x-1|+\frac{5}{8}\log |x+3|+\frac{1}{2}\frac{{(x-1)}^{-1}}{-1}+c$

$=\frac{3}{8}\log |x-1|+\frac{5}{8}\log |x+3|-\frac{1}{2(x-1)}+c$