express the integrand as a sum of partial fractions and evaluate the integral.\n int\frac{24s +…

Explanation:

Step1: Set up partial - fraction decomposition

We assume that $\frac{24s + 24}{(s^{2}+1)(s - 1)^{3}}=\frac{As + B}{s^{2}+1}+\frac{C}{s - 1}+\frac{D}{(s - 1)^{2}}+\frac{E}{(s - 1)^{3}}$. Then $24s+24=(As + B)(s - 1)^{3}+C(s^{2}+1)(s - 1)^{2}+D(s^{2}+1)(s - 1)+E(s^{2}+1)$.

Step2: Find the value of E

Let $s = 1$. Then $24\times1+24=E(1^{2}+1)$, so $48 = 2E$, and $E = 24$.

Step3: Expand and compare coefficients

Expand the right - hand side: $(As + B)(s - 1)^{3}=(As + B)(s^{3}-3s^{2}+3s - 1)=As^{4}-3As^{3}+3As^{2}-As+Bs^{3}-3Bs^{2}+3Bs - B$. $C(s^{2}+1)(s - 1)^{2}=C(s^{2}+1)(s^{2}-2s + 1)=C(s^{4}-2s^{3}+2s^{2}-2s + 1)$. $D(s^{2}+1)(s - 1)=D(s^{3}-s^{2}+s - 1)$. $E(s^{2}+1)=24s^{2}+24$. Combining like terms and comparing the coefficients of the powers of $s$: For the coefficient of $s^{4}$: $0=A + C$. For the coefficient of $s^{3}$: $0=-3A + B-2C+D$. For the coefficient of $s^{2}$: $0=3A-3B + 2C-D+24$. For the coefficient of $s$: $24=-A + 3B-2C+D$. For the constant term: $24=-B + C-D+24$. From $E = 24$ and solving the system of equations, we find $A=-6$, $B = 6$, $C = 6$, $D = 0$. So $\frac{24s + 24}{(s^{2}+1)(s - 1)^{3}}=\frac{-6s + 6}{s^{2}+1}+\frac{6}{s - 1}+\frac{0}{(s - 1)^{2}}+\frac{24}{(s - 1)^{3}}$.

Step4: Integrate term - by - term

$\int\frac{24s + 24}{(s^{2}+1)(s - 1)^{3}}ds=\int\frac{-6s + 6}{s^{2}+1}ds+\int\frac{6}{s - 1}ds+\int\frac{24}{(s - 1)^{3}}ds$. $\int\frac{-6s + 6}{s^{2}+1}ds=-3\int\frac{2s}{s^{2}+1}ds+6\int\frac{1}{s^{2}+1}ds=-3\ln(s^{2}+1)+6\arctan(s)$. $\int\frac{6}{s - 1}ds=6\ln|s - 1|$. $\int\frac{24}{(s - 1)^{3}}ds=24\int(s - 1)^{-3}ds=-12(s - 1)^{-2}+C$. Combining these results, we get $\int\frac{24s + 24}{(s^{2}+1)(s - 1)^{3}}ds=-3\ln(s^{2}+1)+6\arctan(s)+6\ln|s - 1|-\frac{12}{(s - 1)^{2}}+C$.

Answer:

$-3\ln(s^{2}+1)+6\arctan(s)+6\ln|s - 1|-\frac{12}{(s - 1)^{2}}+C$