Question

$\underset{0}{\overset{1}{\int }}x\log \left(1+\frac{x}{2}\right)dx=a+b\log \frac{2}{3}$ , then

• A. $a=\frac{3}{2},b=\frac{3}{2}$
• B. $a=\frac{3}{4},b=\frac{-}{3}4$
• C. $a=\frac{3}{4},b=\frac{3}{2}$
• D. a = b

Answer 1

Answer: (C)

$\underset{0}{\overset{1}{\int }}xlog\left(1+\frac{x}{2}\right)dx$
$={\left|log\left(1+\frac{x}{2}\right)\cdot \frac{x^2}{2}\right|}_0^1-{\int }_0^1\frac{1}{1+\frac{x}{2}}\cdot \frac{1}{2}\cdot \frac{x^2}{2}dx$
$=\frac{1}{2}log\left(\frac{3}{2}\right)-\frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{x^2}{x+2}dx$
$=\frac{1}{2}log\left(\frac{3}{2}\right)-\frac{1}{2}\underset{0}{\overset{1}{\int }}\left(x-2+\frac{4}{x+2}\right)dx$
$=\frac{1}{2}log\left(\frac{3}{2}\right)-\frac{1}{2}\underset{0}{\overset{1}{\int }}\left(x-2+\frac{4}{x+2}\right)dx$
$=\frac{1}{2}log\left(\frac{3}{2}\right)-\frac{1}{2}{\left[\frac{x^2}{2}-2x+4log\left(x+2\right)\right]}_0^1$
$=\frac{1}{2}log\left(\frac{3}{2}\right)-\frac{1}{2}\left[\frac{1}{2}-2+4log3-4log2\right]$
$=\frac{1}{2}log3-\frac{1}{2}log2-\frac{1}{4}+1-2log3+2log2$
$=\frac{3}{2}log2-\frac{3}{2}log3+\frac{3}{4}=\frac{3}{2}log\left(\frac{2}{3}\right)+\frac{3}{4}$
$\therefore a=\frac{3}{4};b=\frac{3}{2}$.