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  • explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 64 maths

explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 64 maths

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Answer: \frac{1}{(\log 5)^{3}} \cdot 5^{5^{5^{x}}}+c

Hint: Use substitution method to solve this integral

Given:  \int 5^{5^{5^{x}}} 5^{5^{x}} 5^{x} d x

Solution:

        \begin{aligned} &\text { let } I=\int 5^{5^{5^{x}}} 5^{5^{x}} 5^{x} d x \\ &\text { Putting } 5^{5^{5^{x}}}=t \end{aligned}

        \Rightarrow\left(5^{5^{5^{x}}} \cdot \log 5.5^{5^{x}} \cdot \log 5.5^{x} \log 5\right) d x=d t

        \begin{aligned} &\Rightarrow 5^{5^{5^{5}}} 5^{5^{x}} 5^{x}(\log 5)^{3} d x=d t \\ &\Rightarrow\left(5^{5^{5^{x}}} 5^{5^{x}} 5^{x}\right) d x=\frac{d t}{(\log 5)^{3}} \text { then } \end{aligned}

        I=\int \frac{d t}{(\log 5)^{3}}=\frac{1}{(\log 5)^{3}} \int 1 d t

            =\frac{1}{(\log 5)^{3}} \int t^{0} d t=\frac{1}{(\log 5)^{3}} \frac{t^{0+1}}{0+1} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

            =\frac{1}{(\log 5)^{3}} \cdot t+c=\frac{1}{(\log 5)^{3}} \cdot 5^{5^{5^{x}}}+c\left[\because 5^{5^{5^{x}}}=t\right]

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