## University Calculus: Early Transcendentals (3rd Edition)

$$\int\frac{2\theta^3+5\theta^2+8\theta+4}{(\theta^2+2\theta+2)^2}d\theta=\ln(\theta^2+2\theta+2)-\tan^{-1}(\theta+1)-\frac{1}{\theta^2+2\theta+2}+C$$
$$I=\int\frac{2\theta^3+5\theta^2+8\theta+4}{(\theta^2+2\theta+2)^2}d\theta$$ 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 $|\theta^2+2\theta+2|=\theta^2+2\theta+2$ $$I=\ln(\theta^2+2\theta+2)-\tan^{-1}(\theta+1)-\frac{1}{\theta^2+2\theta+2}+C$$