Question

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

The Correct Option is C

$\int\limits^{1/2}_{{0}}$$\frac{\ell n\left(1+2x\right)}{1+\left(2x\right)^{2}}dx \,$ Put $2x=tan\,\theta$ $dx=\frac{1}{2}sec^{2}\,\theta\,d\theta$ at $x=0, \theta=0,$ at $x=\frac{1}{2}, \theta=\frac{\pi}{4}$ $I=\int\limits^{\pi/4}_{{0}}$$\frac{\log\left(1+tan\,\theta\right)}{1+tan^{2}\,\theta }. \frac{1}{2}\,sec^{2}\,\theta \,d\theta$ $I=\frac{1}{2} \int\limits^{\pi/4}_{{0}}\log(1\, tan\,\theta)\, d\theta\frac{1}{2}\,I_1$ $I=\int\limits^{\pi/4}_{{0}}$$\log\left[1+tan\left(\frac{\pi}{4}-\theta\right)\right]$ using property $=\int\limits^{\pi/4}_{{0}}$$\log\left[\frac{2}{1+tan\,\theta}\right]=\int\limits^{\pi/4}_{{0}}\log\, 2\,d\theta-\int\limits^{\pi/4}_{{0}}\,\log(1 +\tan\theta)\,d\theta$ $I_{1}=\frac{\pi}{4}\,\log\,2-I_{1}$ $I_{1}=\frac{\pi}{8}\ln2$ $\Rightarrow I=\frac{\pi}{16}\ln2$