Question

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

Answer 1

$$\begin{gathered} \int \frac{x}{x^2+3x+2}dx \\ x=\mathrm{A}\frac{d}{dx}\left(x^2+3x+2\right)+\mathrm{B} \\ x=\mathrm{A}\left(2x+3\right)+\mathrm{B} \\ x=\left(2\mathrm{A}x\right)+3\mathrm{A}+\mathrm{B} \end{gathered}$$

Comparing the Coefficients of like powers of x we get
$$\begin{gathered} 2\mathrm{A}=1\Rightarrow \mathrm{A}=\frac{1}{2} \\ 3\mathrm{A}+\mathrm{B}=0 \\ \frac{3}{2}+\mathrm{B}=0 \\ \mathrm{B}=-\frac{3}{2} \\ x=\frac{1}{2}\left(2x+3\right)-\frac{3}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\int \frac{x}{x^2+3x+2}dx \\ =\int \left[\frac{{\displaystyle \frac{1}{2}}\left(2x+3\right)-{\displaystyle \frac{3}{2}}}{x^2+3x+2}\right]dx \\ =\frac{1}{2}\int \frac{\left(2x+3\right)dx}{x^2+3x+2}-\frac{3}{2}\int \frac{dx}{x^2+3x+2} \\ =\frac{1}{2}\int \frac{\left(2x+3\right)dx}{x^2+3x+2}-\frac{3}{2}\int \frac{dx}{x^2+3x+{\left({\displaystyle \frac{3}{2}}\right)}^2-{\left({\displaystyle \frac{3}{2}}\right)}^2+2} \\ =\frac{1}{2}\int \frac{\left(2x+3\right)dx}{x^2+3x+2}-\frac{3}{2}\int \frac{dx}{{\left(x+{\displaystyle \frac{3}{2}}\right)}^2-{\displaystyle \frac{9}{4}}+2} \\ =\frac{1}{2}\int \frac{\left(2x+3\right)dx}{x^2+3x+2}-\frac{3}{2}\int \frac{dx}{{\left(x+{\displaystyle \frac{3}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2} \\ =\frac{1}{2}\log \left|x^2+3x+2\right|-\frac{3}{2}\times \frac{1}{2\times {\displaystyle \frac{1}{2}}}\log \left|\frac{x+{\displaystyle \frac{3}{2}}-{\displaystyle \frac{1}{2}}}{x+{\displaystyle \frac{3}{2}}+{\displaystyle \frac{1}{2}}}\right|+C \\ =\frac{1}{2}\log \left|x^2+3x+2\right|-\frac{3}{2}\log \left|\frac{x+1}{x+2}\right|+C \end{gathered}$$