Partial Fractions: int (5x + 1)/[x(x - 3)^2] dx

[jeng]

Using partial fractions, how would you integrate the following:

S [(5x+1) dx] / [x(x-3)^2]

 

[soroban]

Hello, jeng!

You didn't tell us exactly where your difficulty is . . .

\(\displaystyle \L\int \frac{5x\,+\,1}{x(x\,-\,3)^2}\,dx\)

The set-up? \(\displaystyle \L\;\frac{5x\,+\,1}{x(x\,-\,3)^2}\;= \;\frac{A}{x}\,+\,\frac{B}{x\,-\,3}\,+\,\frac{C}{(x\,-\,3)^2}\)


Clearing denominators? $\displaystyle \;5x\,+\,1\;=\;A(x\,-\,3)^2\,+\,B(x(x\,-\,3) \,+\,Cx$


Solving for $\displaystyle A,\,B,\,C$ ?

Let $\displaystyle x\,=\,0:\;1\:=\:9A\;\;\Rightarrow\;\;A\,=\,\frac{1}{9}$

Let $\displaystyle x\,=\,3:\;\;16\,=\,3C\;\;\Rightarrow\;\;C\,=\,\frac{16}{3}$

Let $\displaystyle x\,=\,1\;\;6\:=\:4A\,\,2B\,+\,C\;\;\Rightarrow\;\;B\,=\,-\frac{1}{9}$

Hence: \(\displaystyle \L\,\frac{5x\,+\,1}{x(x\,-\,3)^2}\;=\;\frac{\frac{1}{9}}{x}\,+\,\frac{-\frac{1}{9}}{x\,-\,3} \,+\,\frac{\frac{16}{3}}{(x\,-\,3)^2}\)


Integrating ?

\(\displaystyle \L\;\;\frac{1}{9}\int\frac{dx}{x}\,-\,\frac{1}{9}\int\frac{dx}{x\,-\,3} \,+ \,\frac{16}{3}\int\)$\displaystyle (x\,-\,3)^{-2}\,dx$

\(\displaystyle \L\;=\;\frac{1}{9}\ln|x|\,-\,\frac{1}{9}\ln|x\,-\,3|\,-\,\frac{16}{3}(x\,-\,3)^{-1}\,+\,C\)

\(\displaystyle \L\;=\;\frac{1}{9}\ln\left|\frac{x}{x\,-\,3}\right|\,-\,\frac{16}{3(x\,-\,3)} \,+\,C\)