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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.3 Question 15 Maths Textbook Solution.

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Answer: \frac{1}{2} e^{2 x}+2 x-\frac{1}{2} e^{-2 x}+c

Hint: \text { To solve this equation } \int e^{a x} d x \text { formula }

Given: \int\left(e^{x}+\frac{1}{e^{x}}\right)^{2} d x

Solution: =\left(e^{x}+\frac{1}{e^{x}}\right)^{2} \quad\left[\begin{array}{l} (a+b)^{2}=a^{2}+b^{2}+2 a b \\ \int e^{a x} d x=\frac{1}{a} e^{a x}+c, a \neq 0 \end{array}\right]

\begin{aligned} &=\left(e^{x}\right)^{2}+2 \cdot e^{x} \cdot \frac{1}{e^{x}}+\left(\frac{1}{e^{x}}\right)^{2} \\ &=e^{2 x}+2+e^{-2 x} \end{aligned}

\begin{aligned} &\int\left(e^{x}+\frac{1}{e^{x}}\right)^{2} d x=\int\left(e^{2 x}+2+e^{-2 x}\right) d x \\ &=\frac{1}{2} e^{2 x}+2 x-\frac{1}{2} e^{-2 x}+c \\ &=\frac{1}{2}\left(e^{2 x}-e^{-2 x}\right)+2 x+c \end{aligned}

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