Basic integrals (2)

$\begin{aligned} I & = \int_9^2 \frac {\ln(1+2x)}{1+x^2}dx & \small \color{#3D99F6} \text{Let }x = \tan \theta \implies dx = \sec^2 \theta \ d\theta \\ & = \int_0^{\tan^{-1}2} \frac {\ln(1+2\tan \theta)\sec^2 \theta}{1+\tan^2\theta} d\theta \\ & = \int_0^{\tan^{-1}2} \ln(1+2\tan \theta) \ d\theta \\ & = \int_0^{\tan^{-1}2} \ln \left(1 + \frac {2\sin \theta}{\cos \theta}\right) \ d\theta \\ & = \int_0^{\tan^{-1}2} \ln \left(\frac {\cos \theta + 2\sin \theta}{\cos \theta}\right) \ d\theta \\ & = \int_0^{\tan^{-1}2} \ln \left(\frac {\sqrt 5\cos \left(\theta - \tan^{-1}2\right)}{\cos \theta}\right) \ d\theta \\ & = \int_0^{\tan^{-1}2} \ln \sqrt 5 \ d\theta + {\color{#3D99F6} \int_0^{\tan^{-1}2} \ln \left(\cos \left(\theta - \tan^{-1}2\right)\right)\ d\theta} - \int_0^{\tan^{-1}2} \ln(\cos \theta) \ d\theta & \small \color{#3D99F6} \text{By identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \ln \sqrt 5 \theta \ \bigg|_0^{\tan^{-1}2} + {\color{#3D99F6} \int_0^{\tan^{-1}2} \ln (\cos (- \theta))\ d\theta} - \int_0^{\tan^{-1}2} \ln(\cos \theta) \ d\theta \\ & = \ln \sqrt 5 \tan^{-1}2 + \int_0^{\tan^{-1}2} \ln (\cos \theta)\ d\theta - \int_0^{\tan^{-1}2} \ln(\cos \theta) \ d\theta \\ & = \tan^{-1}2 \ln \sqrt 5 \end{aligned}$

$\implies ab = 2 \times 5 = \boxed{10}$

Brief sketch use differentiate under integral with I(a)=ln(ax+1)/x^2+1 Partial fraction calculation gives result