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

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

Answers (1)

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

Solution:

        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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