How is this integral derived?

[opticaltempest]

I need to solve an integral of the form,

\(\displaystyle \L
\int {\frac{u}{{a + bu}}} {\rm }du = {\rm }\frac{1}{{b^2 }}(bu - a\ln (\left| {a + bu} \right|) + C\)

Where a and b are constants...

How is the integral derived? This was an integral in a table of integrals. I'v e been trying to use integration by parts to solve my specific integral but I am not having much luck. Thanks

 

[pka]

Find the derivative of \(\displaystyle \L
\ {\rm }\frac{1}{{b^2 }}(bu - a\ln (\left| {a + bu} \right|) + C\).

If you study the result, then you will see what happens.

 

[soroban]

Hello, opticaltempest!

\(\displaystyle \L\int {\frac{u}{a + bu}\)\(\displaystyle \,du\;=\;\frac{1}{b^2 }\,\left(bu\,-\,a\,\ln|a\,+\,bu|\right)\,+\,C\) . . . where \(\displaystyle a\) and \(\displaystyle b\) are constants.

Note that the rational function is "improper" . . . the top and bottom have the same degree.

They used long division and got: \(\displaystyle \L\:\int\left(\frac{1}{b}\,-\,\frac{a}{b}\cdot\frac{1}{a\,+\,bu}\right)\)\(\displaystyle \,du\)

 

[tkhunny]

You can also look up "Partial Fractions".

 

[opticaltempest]

How do I use partial fraction decomposition on this expression with unknown constants?

m and r represent constants,

\(\displaystyle \L\frac{t}{{m - rt}}\)

\(\displaystyle \L
\frac{t}{{m - rt}} = A + \frac{B}{{m - rt}}\)

\(\displaystyle \L t = A(m - rt) + B{\rm (1)}\)

Is there a way to decompose the fraction without knowing the value of the constants m and r?

I am given a peice of information in the beginning of the problem stating

\(\displaystyle \L
t < \frac{m}{r}\)


How do I go about using long division on

\(\displaystyle \L\frac{t}{{m - rt}}\)

to make it proper?

 

[tkhunny]

What difference does it make if you know or don't know the constants?

Since the denominator is only Linear, this really isn't a job for Partial Fraction decomposition. Long division will do just fine. You still don't need to know the values of any constants. Just drag them along. You have it already.

t = A(m-rt) + B

1 = -rA ==> A = -1/r
0 = Am + B ==> B = m/r

 

[opticaltempest]

I'm still not seeing how to use partial fractions or long division on this expression.

Could someone walk me through the long division process on this rational expression?

I am instantly stumped when trying to apply long division to this expression. I understand how to do polynomial long division on other rational expressions. When looking at this using long division it appears to me to be a proper rational expression.

Where can I reference more information on how to do these?


Side Note: I'm noticing that when using LaTeX code copied from TeXaide (Using "LaTeX 2.09 and Later" format) it will not always display the same here. This is frustrating! Why is LaTeX code being interpreted in different ways? The code is nothing but standards of how to format text! Why is this happening?

 

[pka]

Side Note: I'm noticing that when using LaTeX code copied from TeXaide (Using "LaTeX 2.09 and Later" format) it will not always display the same here. This is frustrating! Why is LaTeX code being interpreted in different ways? The code is nothing but standards of how to format text! Why is this happening?

Isn’t that absolutely madding! But surely by now you understand that the is nothing, absolutely nothing, standard on the web. There are simply different translators.

 

[galactus]

I'm still not seeing how to use partial fractions or long division on this expression.

Could someone walk me through the long division process on this rational expression?

I am instantly stumped when trying to apply long division to this expression. I understand how to do polynomial long division on other rational expressions. When looking at this using long division it appears to me to be a proper rational expression.

Where can I reference more information on how to do these?

This is just the ususal long division. Don't be put-off by the variables. I

know, you're used to seeing integers and it makes it easier to work with.

Let me try to set this up. Hope it works.

First, u/(bu)=1/b

          1/b        
        --------------------
   a+bu|  u
          u+a/b
        -------------------
            -a/b

All there is to it.

We have:

  1             a
-----  -  ----------
  b         b(a+bu)

See?.