If $$\int \frac{\cos\theta}{5 + 7\sin\theta - 2\cos^2\theta}\,d\theta = A\log_e|B(\theta)| + C$$, where $$C$$ is a constant of integration, then $$\frac{B(\theta)}{A}$$ can be:

We have to evaluate the integral

$$\int \dfrac{\cos\theta}{5 + 7\sin\theta - 2\cos^2\theta}\,d\theta.$$

First we put $$t=\sin\theta.$$ Then $$\dfrac{dt}{d\theta}=\cos\theta$$, that is, $$dt=\cos\theta\,d\theta.$$ Therefore

$$\int \dfrac{\cos\theta}{5 + 7\sin\theta - 2\cos^2\theta}\,d\theta \;=\;\int \dfrac{dt}{5 + 7t - 2\cos^2\theta}.$$

Now we replace $$\cos^2\theta$$ in terms of $$t.$$ Using the Pythagorean identity $$\sin^2\theta+\cos^2\theta=1$$ we have $$\cos^2\theta = 1-\sin^2\theta = 1-t^2.$$ Substituting this, the denominator becomes

$$5 + 7t - 2(1 - t^2)=5 + 7t - 2 + 2t^2 = 3 + 7t + 2t^2.$$

Hence the integral turns into

$$\int \dfrac{dt}{2t^2+7t+3}.$$

Next we factor the quadratic $$2t^2+7t+3.$$ Observing that

$$(2t+1)(t+3)=2t^2+6t+t+3=2t^2+7t+3,$$

we write

$$\int \dfrac{dt}{(2t+1)(t+3)}.$$

We decompose the rational function into partial fractions. Let

$$\dfrac{1}{(2t+1)(t+3)}=\dfrac{A}{2t+1}+\dfrac{B}{t+3}.$$

Multiplying both sides by $$(2t+1)(t+3)$$ gives

$$1=A(t+3)+B(2t+1).$$

Expanding and grouping like terms,

$$1=(A+2B)t+(3A+B).$$

Since this equation must hold for all $$t$$, we equate coefficients:

$$A+2B=0,\qquad 3A+B=1.$$

Solving these simultaneously, from $$A+2B=0$$ we have $$A=-2B.$$ Substituting into $$3A+B=1$$ gives $$3(-2B)+B=-6B+B=-5B=1,$$ hence $$B=-\dfrac15.$$ Then $$A=-2B=2\left(\dfrac15\right)=\dfrac25.$$

So

$$\dfrac{1}{(2t+1)(t+3)}=\dfrac{\dfrac25}{2t+1}+\dfrac{-\dfrac15}{t+3} =\dfrac25\cdot\dfrac1{2t+1}-\dfrac15\cdot\dfrac1{t+3}.$$

The integral therefore becomes

$$\int\left(\dfrac25\cdot\dfrac1{2t+1}-\dfrac15\cdot\dfrac1{t+3}\right)dt.$$

We integrate term by term. We first recall the standard formula

$$\int \dfrac{dx}{ax+b}=\dfrac1a \ln|ax+b|.$$

Applying it:

$$\int \dfrac25\cdot\dfrac1{2t+1}\,dt =\dfrac25\cdot\dfrac12\ln|2t+1| =\dfrac15\ln|2t+1|,$$

and

$$\int -\dfrac15\cdot\dfrac1{t+3}\,dt =-\dfrac15\ln|t+3|.$$

Combining, the antiderivative is

$$\dfrac15\ln|2t+1|-\dfrac15\ln|t+3|+C =\dfrac15\ln\left|\dfrac{2t+1}{t+3}\right|+C.$$

We now substitute back $$t=\sin\theta$$ to get

$$\int \dfrac{\cos\theta}{5 + 7\sin\theta - 2\cos^2\theta}\,d\theta =\dfrac15\ln\left|\dfrac{2\sin\theta+1}{\sin\theta+3}\right|+C.$$

This result has the form $$A\log_e|B(\theta)|+C$$ with

$$A=\dfrac15,\qquad B(\theta)=\dfrac{2\sin\theta+1}{\sin\theta+3}.$$

Therefore

$$\dfrac{B(\theta)}{A} =\dfrac{\dfrac{2\sin\theta+1}{\sin\theta+3}}{\dfrac15} =5\cdot\dfrac{2\sin\theta+1}{\sin\theta+3} =\dfrac{5(2\sin\theta+1)}{\sin\theta+3}.$$

Hence, the correct answer is Option D.