Question

$\int \frac{x-3}{x^2+2x-4}dx$

Answer 1

$$\begin{gathered} \int \left(\frac{x-3}{x^2+2x-4}\right)dx \\ x-3=\mathrm{A}\frac{d}{dx}\left(x^2+2x-4\right)+\mathrm{B} \\ x-3=\mathrm{A}\left(2x+2\right)+\mathrm{B} \\ x-3=\left(2\mathrm{A}\right)x+2\mathrm{A}+\mathrm{B} \end{gathered}$$

Comparing Coefficients of like powers of x

$$\begin{gathered} 2\mathrm{A}=1 \\ \mathrm{A}=\frac{1}{2} \\ 2\mathrm{A}+\mathrm{B}=-3 \\ 2\times \frac{1}{2}+\mathrm{B}=-3 \\ \mathrm{B}=-4 \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\int \left(\frac{x-3}{x^2+2x-4}\right)dx \\ =\int \left(\frac{{\displaystyle \frac{1}{2}}\left(2x+2\right)-4}{x^2+2x-4}\right)dx \\ =\frac{1}{2}\int \frac{{\displaystyle \left(2x+2\right)dx}}{\left(x^2+2x-4\right)}-4\int \frac{dx}{x^2+2x+1-1-4} \\ =\frac{1}{2}\int \frac{\left(2x+2\right)dx}{\left(x^2+2x-4\right)}-4\int \frac{dx}{{\left(x+1\right)}^2-{\left({\displaystyle \sqrt{5}}\right)}^2} \\ =\frac{1}{2}\log \left|x^2+2x-4\right|-\frac{4}{2\sqrt{5}}\log \left|\frac{x+1-\sqrt{5}}{x+1+\sqrt{5}}\right|+C \\ =\frac{1}{2}\log \left|x^2+2x-4\right|-\frac{2}{\sqrt{5}}\log \left|\frac{x+1-\sqrt{5}}{x+1+\sqrt{5}}\right|+C \end{gathered}$$