# Integration by Partial Fractions - Long Problem

1. Sep 21, 2008

### RedBarchetta

1. The problem statement, all variables and given/known data
$$\int {\frac{{2s + 2}} {{(s^2 + 1)(s - 1)^3 }}ds}$$

3. The attempt at a solution

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

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

2. Sep 21, 2008

### RedBarchetta

Part 2:

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

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

After solving....

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

3. Sep 21, 2008

### RedBarchetta

Part 3:

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

4. Sep 21, 2008

### RedBarchetta

Part 4:

Alright, now just solving the integral:

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

Here's what the book says:

$$\int {\frac{{2s + 2}} {{(s^2 + 1)(s - 1)^3 }}} = \frac{{ - 1}} {{(s - 1)^2 }} + \frac{1} {{(s - 1)}} + \tan ^{ - 1} s + C$$

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

Last edited: Sep 21, 2008
5. Sep 21, 2008

### 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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook