Question

${\int }_0^1\frac{2x^2+3x+3}{(x-1)(x^2+2x+2)}dx$

Answer 1

${\int }_0^1\frac{2x^2+3x+3}{(x-1)(x^2+2x+2)}.dx$

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

$2x^2+3x+3=A(x^2+2x+2)+(Bx+C)(x-1)$

Put $x=0$,

$3=2A-C.....\left(i\right)$

Put $x=1$, Put $A=\frac{8}{5}$ in eq. $\left(i\right)$

$8=5A\therefore A=\frac{8}{5}C=2A-3=2\times \frac{8}{5}-3$

$=\frac{16-15}{5}$

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

Put $x=-1$,

$2=A+2B-2C......\left(ii\right)$

$2=\frac{8}{5}+2B-\frac{2}{5}$

$2=\frac{6}{5}+2B$ $\therefore B=\frac{2}{5}$

$\therefore \int \frac{A}{x+1}+\frac{Bx+C}{x^2+2x+2}.dx={\int }_0^1\frac{8}{5(x+1)}+\frac{\frac{2}{5}x+\frac{1}{5}}{x^2+2x+2}.dx$

$=\frac{8}{5}{\left[\log \right(x+1\left)\right]}_0^1+{\int }_0^1\frac{2x+1}{5x^2+10x+10}dx$

$=\frac{8}{5}[\log 2-\log 1]+\frac{1}{5}{\int }_0^1\frac{2x+2-2}{x^2+2x+2}.dx+1{\int }_0^1\frac{1}{x^2+2x+2}.dx$

$=\frac{8}{5}\left(\log 2\right)+\frac{1}{5}{\left[\log \right(x^2+2x+2\left)\right]}_0^1-{\int }_0^1\frac{1}{x^2+2x+2}dx$

$=\frac{8}{5}\log 2+1\log 5-15\log 2-I_1$