Question

Evaluate the integral \[ \int \frac{x^2 - 1}{x^2 + 4} \, dx. \]

Hint:

Simplify rational expressions by splitting the numerator to match the denominator, then integrate term by term.

Answer 1

Rewrite the integrand: \[ \frac{x^2 - 1}{x^2 + 4} = \frac{(x^2 + 4) - 5}{x^2 + 4} = 1 - \frac{5}{x^2 + 4}. \] Thus, \[ \int \frac{x^2 - 1}{x^2 + 4} \, dx = \int \left( 1 - \frac{5}{x^2 + 4} \right) dx = \int 1 \, dx - 5 \int \frac{dx}{x^2 + 4}. \] Calculate each integral: \[ \int 1 \, dx = x, \] and \[ \int \frac{dx}{x^2 + 4} = \frac{1}{2} \tan^{-1} \frac{x}{2}. \] Therefore, \[ \int \frac{x^2 - 1}{x^2 + 4} \, dx = x - 5 \times \frac{1}{2} \tan^{-1} \frac{x}{2} + C = x - \frac{5}{2} \tan^{-1} \frac{x}{2} + C. \]
Final answer: \[ \boxed{ x - \frac{5}{2} \tan^{-1} \frac{x}{2} + C. } \]