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

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

Answers (1)

Answer:

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}

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