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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 21 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise Multiple Choice Questions Question 21 Maths Textbook Solution.

Answers (1)

Answer:

\frac{-e^{-x}}{e^{x}+e^{-x}}+C

Given:

\int \frac{2}{\left(e^{x}+e^{-x}\right)^{2}} d x

Hint:

You must know about the \int x^{n} d x

Explanation:

\begin{aligned} &I=\int \frac{2 d x}{\left(e^{x}+e^{-x}\right)^{2}} \\ &I=\int \frac{2 e^{x}}{e^{x}\left(e^{x}+e^{-x}\right)^{2}} d x \\ &\text { Put } e^{x}=t \Rightarrow e^{x} d x=d t \\ &I=\int \frac{2 d t}{t\left(t+\frac{1}{t}\right)^{2}} \\ &I=\int \frac{2 d t}{\frac{t\left(t^{2}+1\right)^{2}}{t^{2}}} \\ &=\int \frac{2 t d t}{\left(t^{2}+1\right)^{2}} \\ &\text { Put } z=t^{2}+1 \Rightarrow d z=2 t d t \end{aligned}

\begin{aligned} &I=\int \frac{d z}{z^{2}} \\ &=\int z^{-2} d z \\ &=\frac{z^{-2+1}}{-2+1}+C \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \\ &=-\frac{1}{z}+C \\ &=\frac{-1}{\left(t^{2}+1\right)}+C \quad\left[\because z=t^{2}+1\right] \\ &=\frac{-1}{e^{2 x}+1}+C \quad\left[\because t=e^{x}\right] \\ &=\frac{-1}{e^{x}\left(e^{x}+\frac{1}{e^{x}}\right)}+C \\ &=\frac{-e^{-x}}{\left(e^{x}+e^{-x}\right)}+C \end{aligned}

Hence I=\frac{-e^{-x}}{e^{x}+e^{-x}}+C

Option (a) is correct

 

Posted by

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