MHB 206.8.5.49 Express the integrand as sum of partial fractions

[karush]

$\tiny{206.8.5.49}$
$\textsf{Express the integrand as sum of partial fractions}$
\begin{align}
&& I_{49}&=\int\frac{30s+30}{(s^2+1)(s-1)^3}\, ds& &(1)& \\
&\textsf{expand}& \\
&& &=\displaystyle
15\int\frac{1}{(s^2+1)}\, ds
-15\int\frac{1}{(s-1)^2}\, ds
+30\int\frac{1}{(s-1)^3}\, ds& &(2)& \\
&\textsf{integrate}&\\
&& &=\displaystyle 15\arctan\left(s\right)
-\frac{-15}{s-1} +\frac{15}{(s-1)^2}& &(3)&\\
&\textsf{simplify}&\\
&& I_{49}&=\displaystyle15\arctan\left(s\right)
+\dfrac{15s-30}{\left(s-1\right)^2}+C& &(4)& \\
\end{align}

$\textsf{the demonator could have expanded to a polynomial?
also some answers had ln in the final answer}$

 

[Prove It]

$\tiny{206.8.5.49}$
$\textsf{Express the integrand as sum of partial fractions}$
\begin{align}
&& I_{49}&=\int\frac{30s+30}{(s^2+1)(s-1)^3}\, ds& &(1)& \\
&\textsf{expand}& \\
&& &=\displaystyle
15\int\frac{1}{(s^2+1)}\, ds
-15\int\frac{1}{(s-1)^2}\, ds
+30\int\frac{1}{(s-1)^3}\, ds& &(2)& \\
&\textsf{integrate}&\\
&& &=\displaystyle 15\arctan\left(s\right)
-\frac{-15}{s-1} +\frac{15}{(s-1)^2}& &(3)&\\
&\textsf{simplify}&\\
&& I_{49}&=\displaystyle15\arctan\left(s\right)
+\dfrac{15s-30}{\left(s-1\right)^2}+C& &(4)& \\
\end{align}

$\textsf{the demonator could have expanded to a polynomial?
also some answers had ln in the final answer}$

The correct partial fraction decomposition you need is $\displaystyle \begin{align*} \frac{A\,s + B}{s^2 +1} + \frac{C}{s - 1} + \frac{D}{\left( s - 1 \right) ^2} + \frac{E}{\left( s - 1 \right) ^3} \end{align*}$.

 

[karush]

The correct partial fraction decomposition you need is $\displaystyle \begin{align*} \frac{A\,s + B}{s^2 +1} + \frac{C}{s - 1} + \frac{D}{\left( s - 1 \right) ^2} + \frac{E}{\left( s - 1 \right) ^3} \end{align*}$.

yes but $A=0$ and $D=0$