Answer
$$\int\frac{2\theta^3+5\theta^2+8\theta+4}{(\theta^2+2\theta+2)^2}d\theta=\ln(\theta^2+2\theta+2)-\tan^{-1}(\theta+1)-\frac{1}{\theta^2+2\theta+2}+C$$
Work Step by Step
$$I=\int\frac{2\theta^3+5\theta^2+8\theta+4}{(\theta^2+2\theta+2)^2}d\theta$$
1) Express the integrand as a sum of partial fractions:
Here we have $(\theta^2+2\theta+2)^2$ both being an irreducible quadratic factor and a repeated linear factor,
$$\frac{2\theta^3+5\theta^2+8\theta+4}{(\theta^2+2\theta+2)^2}=\frac{A\theta+B}{\theta^2+2\theta+2}+\frac{C\theta+D}{(\theta^2+2\theta+2)^2}$$
Clear fractions:
$$(A\theta+B)(\theta^2+2\theta+2)+C\theta+D=2\theta^3+5\theta^2+8\theta+4$$ $$A\theta^3+2A\theta^2+B\theta^2+2A\theta+2B\theta+2B+C\theta+D=2\theta^3+5\theta^2+8\theta+4$$ $$A\theta^3+(2A+B)\theta^2+(2A+2B+C)\theta+(2B+D)=2\theta^3+5\theta^2+8\theta+4$$
Equating coefficients of corresponding powers of $\theta$, we get
$A=2$
$2A+B=5$, so $B=1$
$2A+2B+C=8$, so $C=8-2\times2-2\times1=2$
$2B+D=4$, so $D=2$
Therefore,
$$\frac{2\theta^3+5\theta^2+8\theta+4}{(\theta^2+2\theta+2)^2}=\frac{2\theta+1}{\theta^2+2\theta+2}+\frac{2\theta+2}{(\theta^2+2\theta+2)^2}$$
2) Evaluate the integral:
$$I=\int\frac{2\theta+1}{\theta^2+2\theta+2}d\theta+\int\frac{2\theta+2}{(\theta^2+2\theta+2)^2}d\theta$$ $$I=\int\frac{2\theta+2}{\theta^2+2\theta+2}d\theta-\int\frac{d\theta}{\theta^2+2\theta+2}+\int\frac{1}{(\theta^2+2\theta+2)^2}d(\theta^2+2\theta+2)$$ $$I=\int\frac{1}{\theta^2+2\theta+2}d(\theta^2+2\theta+2)-\int\frac{d(\theta+1)}{(\theta+1)^2+1^2}-\frac{1}{\theta^2+2\theta+2}$$ $$I=\ln|\theta^2+2\theta+2|-\tan^{-1}(\theta+1)-\frac{1}{\theta^2+2\theta+2}+C$$
As $\theta^2+2\theta+2=(\theta+1)^2+1\gt0$ for all $\theta$, we have $|\theta^2+2\theta+2|=\theta^2+2\theta+2$
$$I=\ln(\theta^2+2\theta+2)-\tan^{-1}(\theta+1)-\frac{1}{\theta^2+2\theta+2}+C$$
