using the partial fraction decomposition above, evaluate the following integral.\n int \frac{3y^{3}+28y^{2}+7…

using the partial fraction decomposition above, evaluate the following integral.\n int \frac{3y^{3}+28y^{2}+79y + 118}{(y + 5)^{2}(y + 1)^{2}}dy

using the partial fraction decomposition above, evaluate the following integral.\n int \frac{3y^{3}+28y^{2}+79y + 118}{(y + 5)^{2}(y + 1)^{2}}dy

Answer

Explanation:

Step1: Set up partial - fraction decomposition

Let $\frac{3y^{3}+28y^{2}+79y + 118}{(y + 5)^{2}(y + 1)^{2}}=\frac{A}{y+5}+\frac{B}{(y + 5)^{2}}+\frac{C}{y+1}+\frac{D}{(y + 1)^{2}}$. Then $3y^{3}+28y^{2}+79y + 118=A(y + 5)(y + 1)^{2}+B(y + 1)^{2}+C(y + 1)(y + 5)^{2}+D(y + 5)^{2}$.

Step2: Find the values of A, B, C, D

If $y=-5$, then $3(-5)^{3}+28(-5)^{2}+79(-5)+118=B(-5 + 1)^{2}$, $-375+700-395 + 118=B\times16$, $-152=16B$, $B=- \frac{152}{16}=-\frac{38}{4}=-\frac{19}{2}$. If $y=-1$, then $3(-1)^{3}+28(-1)^{2}+79(-1)+118=D(-1 + 5)^{2}$, $-3 + 28-79+118=D\times16$, $64=16D$, $D = 4$. Expand the right - hand side: $A(y + 5)(y + 1)^{2}+B(y + 1)^{2}+C(y + 1)(y + 5)^{2}+D(y + 5)^{2}=A(y + 5)(y^{2}+2y + 1)+B(y^{2}+2y + 1)+C(y + 1)(y^{2}+10y + 25)+D(y^{2}+10y + 25)=A(y^{3}+2y^{2}+y+5y^{2}+10y + 5)+B(y^{2}+2y + 1)+C(y^{3}+10y^{2}+25y+y^{2}+10y + 25)+D(y^{2}+10y + 25)=A(y^{3}+7y^{2}+11y + 5)+B(y^{2}+2y + 1)+C(y^{3}+11y^{2}+35y + 25)+D(y^{2}+10y + 25)=(A + C)y^{3}+(7A + B+11C+D)y^{2}+(11A + 2B+35C+10D)y+(5A + B+25C+25D)$. Equating the coefficients of $y^{3}$: $A + C=3$. Equating the coefficients of $y^{2}$: $7A + B+11C+D=28$. Substitute $B =-\frac{19}{2}$ and $D = 4$, we get $7A-\frac{19}{2}+11C+4=28$, $7A+11C=28+\frac{19}{2}-4=\frac{56 + 19-8}{2}=\frac{67}{2}$. From $A + C=3$, we have $A=3 - C$. Substitute into $7A+11C=\frac{67}{2}$, $7(3 - C)+11C=\frac{67}{2}$, $21-7C+11C=\frac{67}{2}$, $4C=\frac{67}{2}-21=\frac{67 - 42}{2}=\frac{25}{2}$, $C=\frac{25}{8}$, and $A=3-\frac{25}{8}=\frac{24 - 25}{8}=-\frac{1}{8}$.

Step3: Integrate each term

$\int\frac{3y^{3}+28y^{2}+79y + 118}{(y + 5)^{2}(y + 1)^{2}}dy=\int\left(-\frac{1}{8(y+5)}-\frac{19}{2(y + 5)^{2}}+\frac{25}{8(y+1)}+\frac{4}{(y + 1)^{2}}\right)dy$ $=-\frac{1}{8}\int\frac{1}{y+5}dy-\frac{19}{2}\int\frac{1}{(y + 5)^{2}}dy+\frac{25}{8}\int\frac{1}{y+1}dy+4\int\frac{1}{(y + 1)^{2}}dy$. We know that $\int\frac{1}{u}du=\ln|u|+C$ and $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n\neq - 1$). $=-\frac{1}{8}\ln|y + 5|+\frac{19}{2(y + 5)}+\frac{25}{8}\ln|y + 1|-\frac{4}{y + 1}+C$. $=\frac{1}{8}(25\ln|y + 1|- \ln|y + 5|)+\frac{19}{2(y + 5)}-\frac{4}{y + 1}+C$.

Answer:

$\frac{1}{8}(25\ln|y + 1|-\ln|y + 5|)+\frac{19}{2(y + 5)}-\frac{4}{y + 1}+C$