Question

$\int \frac{1}{x^2+4x-5}dx$

Answer 1

$$\begin{gathered} \int \frac{1}{x^2+4x-5}dx \\ =\int \frac{1}{x^2+4x+4-4-5}dx \\ =\int \frac{1}{x^2+4x+4-3^2}dx \\ =\int \frac{1}{{\left(x+2\right)}^2-3^2}dx \\ =\frac{1}{2\times 3}\ln \left|\frac{x+2-3}{x+2+3}\right|+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \int \frac{\mathit{1}}{x^{\mathit{2}}\mathit{-}{a}^{\mathit{2}}}dx=\frac{1}{2a}\ln \mathit{}\left|\frac{x\mathit{-}a}{x\mathit{+}a}\right|+C\right] \\ =\frac{1}{6}\ln \left|\frac{x-1}{x+5}\right|+C \end{gathered}$$