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

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}

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

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

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