Integration by Partial Fractions - Long Problem

[RedBarchetta]

Homework Statement

<br /> \int {\frac{{2s + 2}}<br /> {{(s^2 + 1)(s - 1)^3 }}ds} <br />

The Attempt at a Solution

This is a long one...First, I split the integrand into partial fractions and find the coefficients:

<br /> \begin{gathered}<br /> \frac{{2s + 2}}<br /> {{(s^2 + 1)(s - 1)^3 }} = \frac{{As + B}}<br /> {{s^2 + 1}} + \frac{C}<br /> {{s - 1}} + \frac{D}<br /> {{(s - 1)^2 }} + \frac{E}<br /> {{(s - 1)^3 }} \hfill \\<br /> 2s + 2 = (As + B)(s - 1)^3 + C(s^2 + 1)(s - 1)^2 + D(s^2 + 1)(s - 1) + E(s^2 + 1) \hfill \\<br /> 2s + 2 = (As + B)(s^3 - 3s^2 + 3s - 1) + C(s^4 - 2s^3 + 2s^2 - 2s + 1) + D(s^3 - s^2 + s - 1) + E(s^2 + 1) \hfill \\ <br /> \end{gathered} <br />

 

[RedBarchetta]

Part 2:

Now find the coefficients. I suppose you can use a calculator if you want to check for these.

<br /> \begin{gathered}<br /> s^4 :A + C = 0 \hfill \\<br /> s^3 :B - 3A - 2C + D = 0 \hfill \\<br /> s^2 :3A - 3B + 2C - D = 0 \hfill \\<br /> s^1 :3B - A - 2C + D = 2 \hfill \\<br /> s^0 :C - B - D + E = 2 \hfill \\ <br /> \end{gathered} <br />

After solving...

<br /> \begin{gathered}<br /> A: - 1/2 \hfill \\<br /> B:1/2 \hfill \\<br /> C:1/2 \hfill \\<br /> D: - 1 \hfill \\<br /> E:1 \hfill \\ <br /> \end{gathered} <br />

 

[RedBarchetta]

Part 3:

<br /> \begin{gathered}<br /> \int {\frac{{2s + 2}}<br /> {{(s^2 + 1)(s - 1)^3 }}} = \int {\left[ {\frac{{As + B}}<br /> {{s^2 + 1}} + \frac{C}<br /> {{s - 1}} + \frac{D}<br /> {{(s - 1)^2 }} + \frac{E}<br /> {{(s - 1)^3 }}} \right]} ds \hfill \\<br /> \int {\frac{{2s + 2}}<br /> {{(s^2 + 1)(s - 1)^3 }}} = \int {\left[ {\frac{{( - 1/2)s + (1/2)}}<br /> {{s^2 + 1}} + \frac{{1/2}}<br /> {{s - 1}} + \frac{{ - 1}}<br /> {{(s - 1)^2 }} + \frac{1}<br /> {{(s - 1)^3 }}} \right]} ds \hfill \\ <br /> \end{gathered} <br />

 

[RedBarchetta]

Part 4:

Alright, now just solving the integral:

<br /> \begin{gathered}<br /> \int {\frac{{2s + 2}}<br /> {{(s^2 + 1)(s - 1)^3 }}} = \frac{1}<br /> {2}\int {\frac{{ds}}<br /> {{s^2 + 1}} - \frac{1}<br /> {2}\int {\frac{{s ds}}<br /> {{s^2 + 1}}} + \int {\frac{{1/2}}<br /> {{s - 1}}ds - } } \int {\frac{{ds}}<br /> {{(s - 1)^2 }} + \int {\frac{{ds}}<br /> {{(s - 1)^3 }}} } \hfill \\<br /> \int {\frac{{2s + 2}}<br /> {{(s^2 + 1)(s - 1)^3 }}} = \frac{1}<br /> {2}\tan ^{ - 1} s - \frac{1}<br /> {4}\ln (s^2 + 1) + \frac{1}<br /> {2}\ln |s - 1| + \frac{1}<br /> {{s - 1}} - \frac{1}<br /> {{2(s - 1)^2 }} + C \hfill \\ <br /> \end{gathered} <br />

Here's what the book says:

<br /> \int {\frac{{2s + 2}}<br /> {{(s^2 + 1)(s - 1)^3 }}} = \frac{{ - 1}}<br /> {{(s - 1)^2 }} + \frac{1}<br /> {{(s - 1)}} + \tan ^{ - 1} s + C<br />

So I'm not quite sure where I went wrong...Any input will be appreciated!

 

[HallsofIvy]

While equating coefficients works it is seldom the simplest way to find the coefficients.
from
2s + 2 = (As + B)(s - 1)^3 + C(s^2 + 1)(s - 1)^2 + D(s^2 + 1)(s - 1) + E(s^2 + 1)
If we let s= 1, so that s-1= 0, we get 2+ 2= 4 = B(0)+ C(0)+ D(0)+ E(2) so E= 2, not 1.

Unfortunately, no other value of s makes everything collapse so easily but letting s= 0, -1, 2, and -2 should give fairly simple equations for A, B, C, and D.