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Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.28 Question 6 Maths Textbook Solution.

Answers (1)

Answer:-

\frac{1}{2} e^{x} \sqrt{e^{2 x}+1}+\frac{1}{2} \log \left\{e^{x}+\sqrt{e^{2 x}+1}\right\}+c

Hint:-

Let  e^{x}=z

Given:-

\int e^{x} \sqrt{e^{2 x}+1} d x

Solution:-

By taking  e^{x}=z

e^{x} d x=z d x

Hence

\begin{aligned} &=\int \sqrt{z^{2}+1} d z \\\\ &=\int \sqrt{(z)^{2}+(1)^{2}} d z \\\\ &=\frac{1}{2} z \sqrt{z^{2}+1}+\frac{1}{2}(1)^{2} \log \left\{z+\sqrt{z^{2}+1}\right\}+c \end{aligned}

 

Using the formula

\begin{aligned} &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \\\\ &=\frac{1}{2} e^{x} \sqrt{e^{2 x}+1}+\frac{1}{2} \log \left\{e^{x}+\sqrt{e^{2 x}+1}\right\}+c \end{aligned}

 

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