lf $\int \frac{9 cos x - sin x}{4 sin x + 5 cos x} d_{X} = A$ Iog $| 4 sin x + 5 cos x | +$ Bx $+ c$ , then $A = \dots \dots \dots ., B = \dots \dots .$ ,

\frac{1}{2}, \frac{- 1}{2}

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1, 1

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- 1, 1

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Solution

The correct option is C $1, 1$
$\int \frac{9 c o s x - s i n x}{4 sin x + 5 c o s x} d x$
$N (x) = λ (D^{1} (x)) + μ (D (x))$
$\Rightarrow 9 c o s x - s i n x = λ (4 c o s x - 5 sin x) + μ (5 c o s x + 4 s i n x)$
$\begin{matrix} 4 λ + 5 μ = 9 - 5 λ + 4 m u = - 1 \end{matrix}} λ = μ = 1$
$N (x) = D^{1} (x) + D (x)$
$\int \frac{N (x)}{D (x)} d x = \int \frac{D^{1} (x) + D (x)}{D (x)} d x \int \frac{D^{1} (x)}{D (x)} + \int \frac{D (x)}{D (x)} d x$
$= l o g | D (x) | + x + c$
$= l o g | 4 s i n x + 5 c o s x | + x + c$
$A = 1; B = 1$

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