Question

Find : $\int \frac{x^2+x+1}{(x^2+1)(x+2)}dx$

Answer 1

By the method of partial fractions,

$\frac{x^2+x+1}{(x^2+1)(x+2)}=\frac{A}{x+2}+\frac{Bx+C}{x^2+1}$

$\Rightarrow x^2+x+1=A(x^2+1)+(Bx+C)(x+2)$

Put $x=0\Rightarrow 1=A+2C$ .........$\left(1\right)$

Put $x=1\Rightarrow 3=2A+3B+3C$

Put $A=1-2C$ from $\left(1\right)$ in $2A+3B+3C=3$ we get

$\Rightarrow 2(1-2C)+3B+3C=3$

$\Rightarrow 2-4C+3B+3C=3$

$\Rightarrow 3B-C=1$ .........$\left(2\right)$

Put $x=-1$

$\Rightarrow 1-1+1=2A+(-B+C)\left(1\right)$

$\Rightarrow 2A-B+C=1$

$\Rightarrow 2(1-2C)-B+C=1$ since $A=1-2C$

$\Rightarrow 2-4C-B+C=1$

$\Rightarrow -3C-B=-1$

$\Rightarrow 3C+B=1$

$\Rightarrow B=1-3C$ .......$\left(3\right)$

Put $\left(3\right)$ in $\left(2\right)$ we get $3(1-3C)-C=1$

$\Rightarrow 3-9C-C=1$ or $10C=2$

$\therefore C=\frac{2}{10}=\frac{1}{5}$

$\Rightarrow B=1-3C=1-\frac{3}{5}=\frac{2}{5}$ from $\left(2\right)$

Put $C=\frac{1}{5}$ in $\left(1\right)$ we get

$A=1-2C=1-\frac{2}{5}=\frac{3}{5}$

$\therefore A=\frac{3}{5},B=\frac{2}{5}$ and $C=\frac{1}{5}$

Now,$\int \frac{x^2+x+1}{(x^2+2)(x+1)}dx$

$=\int \frac{Adx}{x+1}+\int \frac{Bx+C}{x^2+2}dx$

$=\frac{3}{5}\int \frac{dx}{x+1}+\frac{1}{5}\int \frac{2x+1}{x^2+2}dx$

$=\frac{3}{5}\int \frac{dx}{x+1}+\frac{1}{5}\int \frac{2x}{x^2+2}dx+\frac{1}{5}\int \frac{dx}{x^2+{\left(\sqrt{2}\right)}^2}$

$=\frac{3}{5}\log |x+1|+\frac{1}{5}\log |x^2+2|+\frac{1}{5}\times \frac{1}{\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}+c$

$=\frac{3}{5}\log |x+1|+\frac{1}{5}\log |x^2+2|+\frac{1}{5\sqrt{2}}{\tan }^{-1}\frac{x}{\sqrt{2}}+c$