#### Explain Solution R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 37 Maths Textbook Solution.

$a=\frac{-1}{8}, b=\frac{+7}{8}$

Given:

$\int \frac{3 e^{x}-5 e^{-x}}{4 e^{x}+5 e^{-x}} d x=a x+b \log _{e}\left|4 e^{x}+5 e^{-x}\right|+C$

Hint:

Using $\int e^{x} d x$ .

Explanation:

We are given that

$\int \frac{3 e^{x}-5 e^{-x}}{4 e^{x}+5 e^{-x}} d x=a x+b \log _{e}\left|4 e^{x}+5 e^{-x}\right|+C$

Differentiate both sides

$\frac{3 e^{x}-5 e^{-x}}{4 e^{x}+5 e^{-x}}=a+b \cdot \frac{1}{\left(4 e^{x}+5 e^{-x}\right)}\left(4 e^{x}-5 e^{-x}\right)$                                    $\left[\because \int e^{x} d x=e^{x}+C\right]$

\begin{aligned} &\Rightarrow \frac{3 e^{x}-5 e^{-x}}{4 e^{x}+5 e^{-x}}=\frac{a\left(4 e^{x}+5 e^{-x}\right)+b\left(4 e^{x}-5 e^{-x}\right)}{4 e^{x}+5 e^{-x}} \\ &\Rightarrow 3 e^{x}-5 e^{-x}=4 a e^{x}+5 a e^{-x}+4 b e^{x}-5 b e^{-x} \\ &\Rightarrow 3 e^{x}-5 e^{-x}=(4 a+4 b) e^{x}+(5 a-5 b) e^{-x} \end{aligned}

Comparing co eff. of like terms, we get

\begin{aligned} &4 a+4 b=3 ; 5(a-b)=-5 \\ &\Rightarrow a-b=-1 \Rightarrow a=b-1 \\ &\Rightarrow 4(b-1)+4 b=3 \\ &\Rightarrow 8 b-4=3 \\ &\Rightarrow 8 b=7 \Rightarrow b=\frac{7}{8} \\ &\therefore a=\frac{7}{8}-1 \Rightarrow a=\frac{-1}{8} \end{aligned}