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

The integral becomes:

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

Using partial fractions, we can write:

$\dfrac{-1}{(2+t)(3+4t)} = \dfrac{A'}{2+t} + \dfrac{B'}{3+4t}$

$-1 = A'(3+4t) + B'(2+t)$

Putting $t = -2$, we get $-1 = A'(3 - 8) \Rightarrow A' = \dfrac{1}{5}$.

Putting $t = -\dfrac{3}{4}$, we get $-1 = B'\left(2 - \dfrac{3}{4}\right) \Rightarrow -1 = B'\left(\dfrac{5}{4}\right) \Rightarrow B' = -\dfrac{4}{5}$.

So, the integral is:

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

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

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

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 the given expression $A\ln|2+\cos\theta| + B\ln|3+4\cos\theta|$, we get:

$B = -\dfrac{1}{5}$

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