Question

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

Answer 1

$$\begin{gathered} \int \frac{x^2-1}{x^2+4}dx \\ =\int \left(\frac{x^2+4-4-1}{x^2+4}\right)dx \\ =\int \left(\frac{x^2+4}{x^2+4}\right)dx-5\int \frac{dx}{x^2+2^2} \\ =\int dx-5\int \frac{dx}{x^2+2^2} \\ =x-\frac{5}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+C\left[\therefore \int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C\right] \end{gathered}$$