Without calculator, is it possible to do?

$\begin{aligned} I & = \int_0^\frac \pi 2 \frac {7\sin \theta - 4\cos \theta + 3}{2\sin \theta + \cos \theta +2} d\theta & \small \color{#3D99F6} \text{Let }t = \tan \frac \theta 2 \implies dt = \frac {\sec^2 \theta}2 d \theta \\ & = \int_0^1 \frac {2\left(7(2t)-4(1-t^2)+3(1+t^2)\right)}{\left(2(2t)+1-t^2+2(1+t^2)\right)(1+t^2)}dt & \small \color{#3D99F6} \implies \sin \theta = \frac {2t}{1+t^2}, \ \cos \theta = \frac {1-t^2}{1+t^2} \\ & = \int_0^1 \frac {14t^2+28t-2}{(t^2+4t+3)(t^2+1)}dt \\ & = \int_0^1 \frac {14t^2+28t-2}{(t+1)(t+3)(t^2+1)}dt & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = \int_0^1 \left(\frac {6t+4}{t^2+1} - \frac 4{t+1} - \frac 2{t+3} \right) dt \\ & = 3\ln(t^2+1) + 4\tan^{-1} t - 4\ln(t+1) - 2\ln(t+3) \ \bigg|_0^1 \\ & = 3\ln 2 + \pi - 4\ln 2 - 4 \ln 2 - 3 \ln 1 - 0 + 4 \ln 1 + 2 \ln 3 \\ & = \pi - 5 \ln 2 + 2 \ln 3 \\ & \approx \boxed{1.87} \end{aligned}$