Passage: Let ∫( θ\,dθ)/((2+ θ)(3+4 θ))=A |2+ θ|+B |3+4 θ| ....

Let $I = \displaystyle\int\dfrac{\sin\theta\,d\theta}{(2+\cos\theta)(3+4\cos\theta)}$

Substitute $\cos\theta = t$, which gives $-\sin\theta\,d\theta = dt \Rightarrow \sin\theta\,d\theta = -dt$

$I = \displaystyle\int\dfrac{-dt}{(2+t)(3+4t)}$

Using partial fractions:

$\dfrac{-1}{(t+2)(4t+3)} = \dfrac{P}{t+2} + \dfrac{Q}{4t+3}$

$-1 = P(4t+3) + Q(t+2)$

Substituting $t = -2$, we get $-1 = P(-8+3) \Rightarrow P = \dfrac{1}{5}$

Substituting $t = -\dfrac{3}{4}$, we get $-1 = Q\left(-\dfrac{3}{4}+2\right) \Rightarrow Q = -\dfrac{4}{5}$

Therefore, the integral becomes:

$I = \displaystyle\int \left( \dfrac{1/5}{t+2} - \dfrac{4/5}{4t+3} \right) dt$

$I = \dfrac{1}{5}\ln|t+2| - \dfrac{4}{5} \cdot \dfrac{1}{4} \ln|4t+3| + C$

$I = \dfrac{1}{5}\ln|t+2| - \dfrac{1}{5}\ln|4t+3| + C$

Substituting back $t = \cos\theta$:

$I = \dfrac{1}{5}\ln|2+\cos\theta| - \dfrac{1}{5}\ln|3+4\cos\theta| + C$

Comparing this with $A\ln|2+\cos\theta| + B\ln|3+4\cos\theta|$, we get $A = \dfrac{1}{5}$.

Answer: $\dfrac{1}{5}$