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: The substitution $x = \tan\theta$ transforms the arctangent in the denominator into a simple form, converting the integral into one over an angle variable.
Let $I = \int_0^2 \frac{\ln(1+2x)}{1+x^2}dx$. Substitute $x = \tan\theta$ so $dx = \sec^2\theta\,d\theta$; when $x=2$, $\theta = \tan^{-1}(2)$. The integral becomes $\int_0^{\tan^{-1}(2)} \ln(1+2\tan\theta)\,d\theta$, which requires further evaluation using integration properties or special techniques for logarithmic integrals.
Correct Answer: 1,2,3