#### Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 36 Maths Textbook Solution.

$\frac{e^{2}}{2}-e$

Hint: Use indefinite integral formula and put limits to solve this integral

Given:

$\int_{e}^{e^{2}}\left \{ \frac{1}{log\: x}-\frac{1}{\left ( log\: x \right )^{2}} \right \}dx$

Solution:

\begin{aligned} &\int_{e}^{e^{2}}\left\{\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right\} d x=\int_{e}^{e^{2}} \frac{1}{\log x} d x-\int_{e}^{e^{2}} \frac{1}{(\log x)^{2}} d x \\ &=\int_{e}^{e^{2}} \frac{1}{\log x} \cdot 1 d x-\int_{e}^{e^{2}} \frac{1}{(\log x)^{2}} d x \\ &=\left[\frac{1}{\log x} \int 1 d x\right]_{e}^{e^{2}}-\int_{e}^{e^{2}}\left\{\frac{d}{d x}\left(\frac{1}{\log x}\right) \int 1 d x\right\} d x-\int_{e}^{e^{2}} \frac{1}{(\log x)^{2}} d x \\ &\quad\left[\because \int 1 d x=x \& \frac{d}{d x}\left(\frac{1}{x}\right)=\frac{d}{d x}\left(x^{-1}\right)=-1 x^{-1-1}=-1 \cdot x^{-2}=-\frac{1}{x^{2}}\right] \end{aligned}

\begin{aligned} &=\left[\frac{1}{\log x} x\right]_{\theta}^{e^{2}}-\int_{\theta}^{e}\left(-\frac{1}{(\log x)^{2}}\right) \frac{1}{x} x d x-\int_{e}^{e^{2}} \frac{1}{(\log x)^{2}} d x \\ &=\left[\frac{x}{\log x}\right]_{e}^{e^{2}}+\int_{e}^{e^{2}} \frac{1}{(\log x)^{2}} d x-\int_{e}^{e^{2}} \frac{1}{(\log x)^{2}} d x \\ &=\left[\frac{e^{2}}{\log e^{2}}-\frac{e}{\log e}\right] \\ &=\left[\frac{e^{2}}{2 \log e}-\frac{e}{\log e}\right] \; \; \; \; \; \; \; \; \; \; \; \quad[\because \log e=1] \\ &=\left[\frac{e^{2}}{2}-e\right] \end{aligned}