If $\int_0^2 \frac{\ln(1+2x)}{1+x^2} dx = (\tan^{-1}a)(\ln\sqrt{b})$ where $a,b \in \mathbb{N}$, then:
Step-by-Step Solution
Key Concept: Use substitution $x = \tan\theta$ to convert the integral to $\int_0^{\tan^{-1}2} \ln(1+2\tan\theta) d\theta$, then apply King's property (replacing $\theta$ with $\tan^{-1}2 - \theta$) to create a symmetric equation that isolates the integral as $\frac{\tan^{-1}2 \cdot \ln 5}{2}$.
To find $I = \int_0^{\tan^{-1}2} \ln(1 + 2\tan\theta) d\theta$, apply the King property by substituting $\theta \to \tan^{-1}2 - \theta$. This yields $I = \int_0^{\tan^{-1}2} \ln\left(1+2\cdot\frac{2-\tan\theta}{1+2\tan\theta}\right) d\theta = \int_0^{\tan^{-1}2} \ln 5\theta - \int_0^{\tan^{-1}2} \ln(1+2\tan\theta) d\theta$. Solving gives $2I = \tan^{-1}1 \cdot \ln 5$, so $I = \frac{1}{2}\tan^{-1}2 \cdot \ln 5 = \tan^{-1}2 \cdot \ln\sqrt{5}$. Since $a^2 + b^2 = 4 + 25 = 29$ with $a=2, b=5$.
Correct Answer: 1,2,3