#### Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 2 maths textbook solution.

$\frac{\log 2}{\log 2e}$

Hint: We use indefinite integral formula then put limits to solve this integral.

Given:  $\int_{1}^{2}\frac{1}{x(1+\log x)^2}dx$

Solution:  $\int_{1}^{2}\frac{1}{x(1+\log x)^2}dx$

Put $1+\log x=t$

$\frac{1}{x}dx=dt$

$dx=xdt$

When x=1  then t=1  and when x=2  then t=1+log2

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

\begin{aligned} &=\left[\frac{t^{-1}}{-1}\right]_{1}^{1+\log 2}=-\left[\frac{1}{t}\right]_{1}^{1+\log 2} \\ &=-\left[\frac{1}{1+\log 2}-1\right]=-\left[\frac{1-1-\log 2}{1+\log 2}\right] \\ &=-\left[\frac{-\log 2}{1+\log 2}\right]=\frac{\log 2}{1+\log 2} \quad[1=\log e] \end{aligned}

\begin{aligned} &=\frac{\log 2}{\log e+\log 2} \\ &=\frac{\log 2}{\log 2 e} \end{aligned}