MCQ
$\int\limits_0^{\frac{1}{2}} {\frac{{\ln \left( {1 + 2x} \right)}}{{1 + 4{x^2}}}} \,dx$ =
- A$\frac{{\pi \ln 2}}{8}$
- B$\frac{{\pi \ln 2}}{4}$
- C$\frac{{\pi \ln 2}}{32}$
- ✓$\frac{{\pi \ln 2}}{16}$
$I = \frac{1}{2}\int\limits_0^{\pi /4} {\ln \left( {1 + \tan \theta } \right)} d\theta $
$\therefore I = \frac{\pi }{{16}}\ln 2$
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