The integral ∫ _0^ 1 2 ln(1+2x) 1+4x^2dx equals

We have, $\displaystyle I=\int _{0}^{\frac{1}{2}}\frac{\mathrm{ln}(1+2x)}{1+4{x}^{2}}dx$I=∫021​​1+4x2ln(1+2x)​dx

Let, $2x=\mathrm{tan}\theta \Rightarrow dx=\frac{1}{2}se{c}^{2}\theta d\theta$2x=tanθ⇒dx=21​sec2θdθ

$\displaystyle \Rightarrow I=\int _{0}^{\frac{\pi }{4}}\frac{\mathrm{ln}(1+\mathrm{tan}\theta )}{se{c}^{2}\theta }\cdot \frac{1}{2}\cdot se{c}^{2}\theta d\theta$⇒I=∫04π​​sec2θln(1+tanθ)​⋅21​⋅sec2θdθ $(\because 1+{\mathrm{tan}}^{2}\theta ={\mathrm{sec}}^{2}\theta )$(∵1+tan2θ=sec2θ)

$$=\frac{1}{2}\int _{0}^{\frac{\pi }{4}}ln(1+\mathrm{tan}\theta )d\theta$$=21​∫04π​​ln(1+tanθ)dθ

$$=\frac{1}{2}\int _{0}^{\frac{\pi }{4}}ln(1+\mathrm{tan}(\frac{\pi }{4}-\theta ))d\theta$$=21​∫04π​​ln(1+tan(4π​−θ))dθ

$\displaystyle (\because {\int }_{0}^{a}f(x)dx={\int }_{0}^{a}f(a-x)dx)$(∵∫0a​f(x)dx=∫0a​f(a−x)dx)$\displaystyle =\frac{1}{2}\int _{0}^{\frac{\pi }{4}}ln(1+\frac{1-\mathrm{tan}\theta }{1+\mathrm{tan}\theta })d\theta$=21​∫04π​​ln(1+1+tanθ1−tanθ​)dθ

$\displaystyle =\frac{1}{2}\int _{0}^{\frac{\pi }{4}}ln(\frac{2}{1+\mathrm{tan}\theta })d\theta$=21​∫04π​​ln(1+tanθ2​)dθ$\displaystyle =\frac{1}{2}[(\mathrm{ln}2)\int _{0}^{\frac{\pi }{4}}d\theta -\int _{0}^{\frac{\pi }{4}}\mathrm{ln}(1+\mathrm{tan}\theta )d\theta ]$=21​[(ln2)∫04π​​dθ−∫04π​​ln(1+tanθ)dθ]

$$\Rightarrow 2I=(\mathrm{ln}2)\frac{\pi }{4}-2I$$⇒2I=(ln2)4π​−2I

$$\Rightarrow I=\frac{\pi }{16}(\mathrm{ln}2)$$⇒I=16π​(ln2)