Question

The integral ${\int }_{{}_0}^{{}^{\frac{1}{2}}}\frac{ln\left(1+2x\right)}{1+4x^2}dx,$ equals :

• A. $\frac{\pi }{4}$ ln 2
• B. $\frac{\pi }{8}$ ln 2
• C. $\frac{\pi }{16}$ ln 2
• D. $\frac{\pi }{32}$ ln 2

Answer 1

Answer: (C)

$\underset{0}{\overset{1/2}{\int }}$$\frac{\ell n\left(1+2x\right)}{1+{\left(2x\right)}^2}dx$ Put $2x=tan\theta$
$dx=\frac{1}{2}se{c}^2\theta d\theta$
at $x=0,\theta =0,$ at $x=\frac{1}{2},\theta =\frac{\pi }{4}$
$I=\underset{0}{\overset{\pi /4}{\int }}$$\frac{\log \left(1+tan\theta \right)}{1+ta{n}^2\theta }.\frac{1}{2}se{c}^2\theta d\theta$
$I=\frac{1}{2}\underset{0}{\overset{\pi /4}{\int }}\log (1tan\theta )d\theta \frac{1}{2}{I}_1$
$I=\underset{0}{\overset{\pi /4}{\int }}$$\log \left[1+tan\left(\frac{\pi }{4}-\theta \right)\right]$ using property
$=\underset{0}{\overset{\pi /4}{\int }}$$\log \left[\frac{2}{1+tan\theta }\right]=\underset{0}{\overset{\pi /4}{\int }}\log 2d\theta -\underset{0}{\overset{\pi /4}{\int }}\log (1+\tan \theta )d\theta$
$I_1=\frac{\pi }{4}\log 2-I_1$
$I_1=\frac{\pi }{8}\ln 2$
$\Rightarrow I=\frac{\pi }{16}\ln 2$