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provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18 .9 question 54

Answers (1)

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

Hint: Use substitution method to solve this integral.

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

Solution:

        \text { Let } I=\int \frac{e^{m \tan ^{-1} x}}{1+x^{2}} d x

        \begin{aligned} &\text { Put } m \tan ^{-1} x=t \Rightarrow m \frac{1}{1+x^{2}} d x=d t \\ &\Rightarrow d x=\frac{1+x^{2}}{m} d t \end{aligned}

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

        =\frac{1}{m} \int e^{t} d t=\frac{1}{m}\left(e^{t}\right)+c \quad\left[\because \int e^{x} d x=e^{x}+c\right]

        =\frac{1}{m} e^{m \tan ^{-1} x}+c \quad\left[\because t=m \tan ^{-1} x\right]

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