Integral #2

$\begin{aligned} I & = \int_1^{-1} \frac {x^2-3}{x^2+2x+5} dx \\ & = - \int_{-1}^1 \frac {x^2-3}{(x+1)^2+4} dx & \small \blue{\text{Let }x = u-1 \implies dx = du} \\ & = - \int_0^2 \frac {u^2 - 2u -2}{u^2 + 4} du \\ & = - \int_0^2 \frac {u^2 + 4 - 2u -6}{u^2 + 4} du \\ & = - \int_0^2 du + \int_0^2 \frac {2u}{u^2 + 4} du + \blue{\frac 64 \int_0^2 \frac 1{\frac {u^2}4+1} du} & \small \blue{\text{Let }t = \frac u2 \implies dt = \frac {du}2} \\ & = -u \ + \ln (u^2+4)\ \bigg|_0^2 + \blue{3 \int_0^1 \frac 1{t^2+1} dt} \\ & = -2+0 + \ln 8 - \ln 4 + 3 \tan^{-1} t \ \bigg|_0^1 \\ & = -2 + \ln 2 + \frac 34 \pi \end{aligned}$

Therefore $a+b+c = -2+2 + \frac 34 = \frac 34 = \boxed{0.75}$ .

$\int_{1}^{-1}\frac{x^{2} - 3}{x^{2} + 2x + 5}dx = -\int_{-1}^{1}\frac{x^{2} - 3}{x^{2} + 2x + 5}dx = \int_{-1}^{1} (-1 + \frac{2x + 8}{x^{2} + 2x + 5})dx = \int_{-1}^{1} (-1 + \frac{2(x + 1)}{(x + 1)^{2} + 4} + \frac{6}{(x + 1)^{2} + 4})dx = [ -x + \ln((x+ 1)^{2} + 4) + 3\tan^{-1}(\frac{x + 1}{2}) ]_{-1}^{1} = -1 + \ln(8) + 3\tan^{-1}(1) - \ln(4) - 1$

Upon simplification:

$\int_{1}^{-1}\frac{x^{2} - 3}{x^{2} + 2x + 5}dx = -2 + \ln(2) + \frac{3\pi}{4}$

Thus, $a = -2, b = 2, c = \frac{3}{4}$ . $a + b + c = \boxed{\frac{3}{4}}$