Question

Evaluate : ${\int }_3^8\frac{2-3x}{x\sqrt{1+x}}dx$

Answer 1

$\begin{array}{c}{\int }_3^8\frac{2-3x}{x\sqrt{1+x}}dx\tan x1+x=t^{2}dx=2+dt\\ {\int }_2^3\frac{5-3r^2}{\left(t^2-1\right)t}2tdt\\ {\int }_2^3\frac{2}{t^2-1}dt-{\int }_2^{3}1dt\\ 4\times \frac{1}{2}\ln {\left[\frac{t-1}{t+1}\right]}_2^3-6\times 1\\ 2\ln \left[\frac{1}{2}/1/2\right]-6\\ =5.2\end{array}$