Question$\int \frac{4 u-6}{u^{2}+6 u+18} d u$

Studdy Solution
Combine the results from the two integrals:
$\int \frac{4u - 6}{u^2 + 6u + 18} \, du = 2 \ln |u^2 + 6u + 18| + 6 \tan^{-1} \left(\frac{u + 3}{3}\right) + C$
where $C = C_1 + C_2$ is the constant of integration.
The solution to the integral is:
$2 \ln |u^2 + 6u + 18| + 6 \tan^{-1} \left(\frac{u + 3}{3}\right) + C$