#### Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 26 Maths Textbook Solution.

Answer:  $\frac{1}{2}\log 6$

Given:  $\int_{1}^{2} \frac{(x+3)}{x(x+2)} d x$

Hint: Do integration by parts

Solution:  $\int_{1}^{2} \frac{(x+3)}{x(x+2)} d x$

\begin{aligned} &=\int_{1}^{2} \frac{(x+2+1)}{x(x+2)} d x \\ & \end{aligned}

$=\int_{1}^{2} \frac{1}{x} d x+\int_{1}^{2}\left(\frac{1}{x(x+2)}\right) d x \\$

$=\int_{1}^{2} \frac{1}{x} d x+\frac{1}{2} \int_{1}^{2}\left(\frac{x+2-x}{x(x+2)}\right) d x$

\begin{aligned} &=\int_{1}^{2} \frac{1}{x} d x+\frac{1}{2} \int_{1}^{2} \frac{1}{x} d x-\frac{1}{2} \int_{1}^{2} \frac{1}{x+2} d x \\ & \end{aligned}

$=\frac{3}{2} \int_{1}^{2} \frac{1}{x} d x-\frac{1}{2} \int_{1}^{2} \frac{1}{x+2} d x$

\begin{aligned} &=\frac{3}{2}(\log x)_{1}^{2}-\frac{1}{2}[\log (x+2)]_{1}^{2} \\ & \end{aligned}

$=\frac{3}{2} \log 2-\frac{1}{2} \log 4+\frac{1}{2} \log 3 \\$

$=\frac{3}{2} \log 2-\log 2+\frac{1}{2} \log 3$

\begin{aligned} &=\frac{1}{2} \log 2+\frac{1}{2} \log 3=\frac{1}{2}(\log 2+\log 3) \\ & \end{aligned}

$=\frac{1}{2} \log 6(\therefore \log a+\log b=\log a b)$