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#### Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.24 question 1 sub question (v) maths textbook solution

Answer: $2 x+\log |2 \cos x+\sin x+3|+C$

Hint: Using formula, put numerator = λ(denominator)+µ(derivative of denominator)

Given: $\int \frac{5 \cos x+6}{2 \cos x+\sin x+3} d x$

Explanation:

$\text { Let, } 1=\int \frac{5 \cos x+6}{2 \cos x+\sin x+3} d x$

$\text { Let, } 5 \cos x+6=A(2 \cos x+\sin x+3)+B(-2 \sin x+\cos x)$        $............(i)$

$\Rightarrow 5 \cos x+6=(A-2 B) \operatorname{Sin} x+(2 A+B) \operatorname{Cos} x+3 A$

Comparing co-efficient of like terms,

\begin{aligned} &\mathrm{A}-2 \mathrm{~B}=0\quad \ldots \text { (ii) }\\ &2 \mathrm{~A}+\mathrm{B}=5 \quad \ldots \text { (iii) }\\ &3 A=6 \; \; \; \; \; \; \; \; \quad \ldots \text { (iv) } \end{aligned}

From eqn (iv)   A = 2

Put value of A in equation (ii)

$2-2B = 0 \Rightarrow B =1$

We get B = 1 and A = 2

By putting values of A, B and C in equation $(ii)$

$\therefore I=\int\left[\frac{2(2 \operatorname{cos} x+\sin x+3)+(-2 \sin x+\cos x)}{2 \cos x+\sin x+3}\right] d x$

$=2 \int \mathrm{dx}+\int \frac{(-2 \sin x+\cos x)}{(2 \cos x+\sin x+3)} \mathrm{d} \mathrm{x}$

Putting $2 \operatorname{cos} x+\sin x+3=t$

$\Rightarrow(-2 \sin x+\cos x) d x=d t$

$\therefore I=2 \int \mathrm{dx}+\int \frac{1}{\mathrm{t}} \mathrm{d} \mathrm{t}$

\begin{aligned} &=2 x+\log |t|+C \\ &\text { Put } t=2 \cos x+\sin x+3 \\ &\therefore I=2 x+\log |2 \cos x+\sin x+3|+c \end{aligned}