Question

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

Answer 1

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

$I=\frac{x^2+x+1}{(x^2+1)(x+2)}=\frac{Ax+B}{x^2+1}+\frac{c}{x+2}$

$x^2+x+1=(Ax+B)(x+2)+c(x^2+1)$

$x^2+x+1=(A+C)x^2+(2A+B)x+(2B+C)$

$A+C=1$ ...(i)

$2A+B=1$ .... (ii)

$2B+C=1$ .... (iii) $\Rightarrow B=\left(\frac{1-C}{2}\right)\to$ substitute in equation (ii)

$2A+\left(\frac{1-c}{2}\right)=1$

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

$+2A+2C=+2$

(-) (-) (-)

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$-\frac{C}{2}-2C=\frac{1}{2}-2$

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$\frac{-5C}{2}=\frac{-3}{2}\Rightarrow C=\frac{3}{5}$

$A+\frac{3}{5}=1A=\frac{2}{5};B=\frac{1}{5}$

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

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

$1+x^2=t$

$2xdx=dt$

$I=\frac{2}{5}\int \frac{x}{t}\times \frac{dt}{2x}+\frac{1}{5}ta{n}^{-1}\left(x\right)+\frac{3}{5}ln|x+2|+C$

$I=\frac{1}{5}ln|t|+\frac{1}{5}ta{n}^{-1}\left(x\right)+\frac{3}{5}ln|x+2|+c$

$I=\frac{1}{5}ln|1+x^2|+\frac{1}{5}ta{n}^{-1}\left(x\right)+\frac{3}{5}ln|x+2|+C$