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Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 66

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Answer: 2 \sqrt{e^{x}-1}-2 \tan ^{-1}\left(\sqrt{e^{x}-1}\right)+c

Hint: Use substitution method to solve this integral

Given: \int \sqrt{e^{x}-1} \; d x


        \begin{aligned} &\operatorname{let} I=\int \sqrt{e^{x}-1} d x \\ &\text { Putting } \mathrm{e}^{x}-1=t^{2} \Rightarrow e^{x} d x=2 t d t \Rightarrow d x=\frac{2 t \cdot d t}{e^{x}} \text { then } \end{aligned}

        I=\int \sqrt{t^{2}} \frac{2 t \; d t}{e^{x}}

            =2 \int \frac{t \cdot t}{t^{2}+1} d t \quad\left[\because e^{x}-1=t^{2} \Rightarrow t^{2}+1=e^{x}\right]

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

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

            =2 \int 1 d t-2 \int \frac{1}{t^{2}+1} d t=2 \int t^{0} d t-2 \int \frac{1}{t^{2}+1} d t

            =2 \frac{t^{0+1}}{0+1}-2 \tan ^{-1}(t)+c                        \left[\begin{array}{l} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ \int \frac{1}{1+x^{2}} d x=\tan ^{-1} x+c \end{array}\right]

            \begin{aligned} &=2 t-2 \tan ^{-1}(t)+c \\ &=2 \sqrt{e^{x}-1}-2 \tan ^{-1}\left(\sqrt{e^{x}-1}\right)+\mathrm{c} \end{aligned}        \left[\because t^{2}=e^{x}-1 \Rightarrow t=\sqrt{e^{x}-1}\right]


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