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

$\frac{1}{6} \ln \left| \frac{x + 1}{5 - x} \right| + C$

The correct answer is Option (1) → $\frac{1}{6} \ln \left| \frac{x + 1}{5 - x} \right| + C$

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

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

$= \int \frac{dx}{(3)^2 - (x - 2)^2}$

$= \frac{1}{2 \times 3} \log \left| \frac{3 + (x - 2)}{3 - (x - 2)} \right| + C$

[using $\int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \log \left| \frac{a + x}{a - x} \right| + C$]

$= \frac{1}{6} \log \left| \frac{x + 1}{5 - x} \right| + C$