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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 49 maths textbook solution

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Answer:\frac{5^{x+\tan ^{-1} x}}{\log 5}+C

Hint: Use substitution method to solve this integral.

Given:   \int 5^{x+\tan ^{-1} x} \cdot\left(\frac{x^{2}+2}{1+x^{2}}\right) d x


        \begin{aligned} &\text { Let } I=\int 5^{x+\tan ^{-1} x} \cdot\left(\frac{x^{2}+2}{1+x^{2}}\right) d x \\ &\text { Put } x+\tan ^{-1} x=t \\ &\Rightarrow\left(1+\frac{1}{1+x^{2}}\right) d x=d t \Rightarrow\left(\frac{\left(1+x^{2}\right)+1}{1+x^{2}}\right) d x=d t \\ &\Rightarrow\left(\frac{1+x^{2}+1}{1+x^{2}}\right) d x=d t \Rightarrow\left(\frac{2+x^{2}}{1+x^{2}}\right) d x=d t \text { then } \end{aligned}

        I=\int 5^{t} d t=\frac{5^{t}}{\log 5}+c \quad\left[\because \int a^{x} d x=\frac{a^{x}}{\log a}+c\right]

            =\frac{5^{x+\tan ^{-1} x}}{\log 5}+c \quad\left[\because t=x+\tan ^{-1} x\right]

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