Question

$\int \frac{2x-3}{x^2+6x+13}dx$

Answer 1

$$\begin{gathered} \int \frac{\left(2x-3\right)dx}{x^2+6x+13} \\ 2x-3=\mathrm{A}\frac{d}{dx}\left(x^2+6x+13\right)+\mathrm{B} \\ 2x-3=\mathrm{A}\left(2x+6\right)+\mathrm{B} \\ 2x-3=\left(2\mathrm{A}\right)x+6\mathrm{A}+\mathrm{B} \end{gathered}$$

Comparing Coefficients of like powers of x
$$\begin{gathered} 2\mathrm{A}=2 \\ \mathrm{A}=1 \\ 6\mathrm{A}+\mathrm{B}=-3 \\ 6+\mathrm{B}=-3 \\ \mathrm{B}=-9 \\ \therefore 2x-3=1\left(2x+6\right)-9 \end{gathered}$$

$$\begin{gathered} \therefore \int \frac{\left(2x-3\right)}{x^2+6x+13}dx \\ =\int \left(\frac{{\displaystyle 2x+6-9}}{x^2+6x+13}\right)dx \\ =\int \left(\frac{2x+6}{x^2+6x+13}\right)dx-\int \frac{9dx}{x^2+6x+13} \\ =\int \frac{\left(2x+6\right)dx}{x^2+6x+13}-9\int \frac{dx}{x^2+6x+3^2-3^2+13} \\ =\int \frac{\left(2x+6\right)dx}{x^2+6x+13}-9\int \frac{dx}{{\left(x+3\right)}^2+2^2} \\ =\log \left|x^2+6x+13\right|-9\times \frac{1}{2}{\tan }^{-1}\left(\frac{x+3}{2}\right)+C \\ =\log \left|x^2+6x+13\right|-\frac{9}{2}{\tan }^{-1}\left(\frac{x+3}{2}\right)+C \end{gathered}$$