Let ∫ sinθ dθ / [(2+cosθ)(3+4cosθ)] = A ln|2+cosθ| + B ln|3+4cosθ|. What is the value of B?

Continuing from the partial fractions setup, integrating -∫ du/[(2+u)(3+4u)] with u = cosθ. With A' = -1/5 and B' = 4/5 for the partial fractions: -∫[(-1/5)/(2+u) + (4/5)/(3+4u)] du = (1/5)ln|2+u| - (1/5)ln|3+4u|. So A = 1/5 and B... rechecking: the coefficient of ln|3+4u| comes from -(4/5)·(1/4) = -1/5. Wait - so B = 1/5.