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Question
\[\int\frac{x^2 - 1}{x^2 + 4} dx\]
Sum
Solution
\[\int\frac{x^2 - 1}{x^2 + 4}dx \]
\[ = \int\left( \frac{x^2 + 4 - 4 - 1}{x^2 + 4} \right)dx \]
\[ = \int\left( \frac{x^2 + 4}{x^2 + 4} \right)dx - 5\int\frac{dx}{x^2 + 2^2}\]
\[ = \int dx - 5\int\frac{dx}{x^2 + 2^2}\]
\[ = x - \frac{5}{2} \tan^{- 1} \left( \frac{x}{2} \right) + C \left[ \therefore \int\frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{- 1} \left( \frac{x}{a} \right) + C \right]\]
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