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explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 60 maths

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Answer: 1+e^{x}-\log \left|1+e^{x}\right|+c

Hint: Use substitution method to solve this integral

Given: \int \frac{e^{2 x}}{1+e^{x}} d x


        I=\int \frac{e^{2 x}}{1+e^{x}} d x

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

                        =\int \frac{e^{x}}{t} d t=\int \frac{t-1}{t} d t \quad\left[\because 1+e^{x} \Rightarrow e^{x}=t-1\right]

                        \begin{aligned} &=\int\left(\frac{t}{t}-\frac{1}{t}\right) d t=\int\left(1-\frac{1}{t}\right) d t \\ &=\int 1 . d t-\int \frac{1}{t} d t=\int t^{0} d t-\int \frac{1}{t} d t \end{aligned}

                        =\frac{t^{0+1}}{0+1}-\log |t|+c                        \left[\begin{array}{l} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ \int \frac{1}{x} d x=\log |x|+c \end{array}\right]

                        \begin{aligned} &=t-\log |t|+c \\ &=1+e^{x}-\log \left|1+e^{x}\right|+c\; \; \; \; \; \; \; \; \; \left[\because t=1+e^{x}\right] \end{aligned}

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