# Integration of dx/[(1-2x)^2*(3-x)]

• Mar 17th 2009, 12:03 AM
gnameni
Integration of dx/[(1-2x)^2*(3-x)]
I have attempted to solve this integration problem by parts but I have not been able to arrive at the final answer. Can anyone help?

Problem: integration of dx/[(1-2x)^2*(3-x)]

• Mar 17th 2009, 12:21 AM
Quote:

Originally Posted by gnameni
I have attempted to solve this integration problem by parts but I have not been able to arrive at the final answer. Can anyone help?

Problem: integration of dx/[(1-2x)^2*(3-x)]

I would say you should prefer intigration by parts

$\int{\frac{dx}{(1-2x)^2 (3-x)}}$

$= \frac{1}{-4}\int{\frac{dx}{(x-1/2)^2(x-3)}}$

Now $\frac{1}{(x-1/2)^2(x-3)} = \frac{A}{x-1/2)^2} + \frac{B}{x-3}$

Find A And B
and then integrate

$\frac{1}{-4}[\int{ \frac{A}{(x-1/2)^2}} +\int{ \frac{B}{x-3}}]$
• Mar 17th 2009, 01:14 AM
gnameni
I do not think that
dx/(1-2x)^2 equals -1/4(dx/(x-1/2)^2)
• Mar 17th 2009, 01:20 AM
Quote:

Originally Posted by gnameni
I do not think that
dx/(1-2x)^2 equals -1/4(dx/(x-1/2)^2)

Neither do I think so (if you are talking about negative sign )
• Mar 17th 2009, 01:30 AM
gnameni
Yes- but that is what you wrote.

I think that the correct expression for the integration of dx/[(1-2x)^2*(3-x)]

is -1/2*integral(dx/[(x-1/2)^2*(x-3)])

How would one go about solving: integral of A/(X-1/2)^2 + integral of B/(x-3)?
• Mar 17th 2009, 01:53 AM
Quote:

Originally Posted by gnameni
Yes- but that is what you wrote.

I think that the correct expression for the integration of dx/[(1-2x)^2*(3-x)]

is -1/2*integral(dx/[(x-1/2)^2*(x-3)])

If you are trying to say that

$
(1-2x)^2(3-x) = -2(x-\frac{1}{2})^2 (x-3)$

than I am sorry sun will not rise from left(Giggle),try again
(Music) Twinkle Twinkle Little Star..

Quote:

How would one go about solving: integral of A/(X-1/2)^2 + integral of B/(x-3)?
Before solving question you have asked you must have solved these types of question

but just for an example I will give A= 1 & B = 1

$\int{\frac{dx}{(x-1/2)^2}}$

Put (x-1/2) = t

dx =dt

Hence integral becomes

$\int{\frac{dt}{(t)^2}}$

$= \frac{-1}{t}+c= \frac{-1}{x-1/2} +c$

doing the same for second integral

$\int{\frac{dx}{x-3}} = ln|x-3| +c$
• Mar 17th 2009, 05:21 AM
Soroban
Hllo, gnameni!

Let's take it from the top . . .

Quote:

$\int \frac{dx}{(1-2x)^2(3-x)}$

$\text{Answer: }\;\frac{1}{5(1-2x)} + \frac{1}{25}\ln\left[\frac{2x-1}{x-3}\right]$
Partial Fractions . . .

We have: . $\frac{1}{(1-2x)^2(3-x)} \;=\;\frac{A}{1-2x} + \frac{B}{(1-2x)^2} + \frac{C}{3-x}$

Then: . $1 \;=\;A(1-2x)(3-x) + B(3-x) + C(1-2x)^2$

Let $x = \tfrac{1}{2}\!:\quad 1 \;=\;A(0) + B(\tfrac{5}{2}) + C(0) \quad\Rightarrow\quad \boxed{B = \tfrac{2}{5}}$ .[1]

Let $x = 3\!:\quad 1 \;=\;A(0) + B(0) + C(25) \quad\Rightarrow\quad\boxed{ C = \tfrac{1}{25}}$ .[2]

Let $x = 0\!:\quad 1 \;=\;A(3) + B(3) + C(1) \quad\Rightarrow\quad 3A + 3B + C \;=\;1$
. . Substitute [1] and [2]: . $3A + \tfrac{6}{5} + \tfrac{1}{25} \:=\:1 \quad\Rightarrow\quad\boxed{ A = \text{-}\tfrac{2}{25}}$

Hence, we have: . $\frac{\text{-}\frac{2}{25}}{1-2x} + \frac{\frac{2}{5}}{(1-2x)^2} + \frac{\frac{1}{25}}{3-x}$

Integrate: . $-\frac{2}{25}\int\frac{dx}{1-2x} + \frac{2}{5}\int(1-2x)^{-2}dx + \frac{1}{25}\int\frac{dx}{3-x}$

. . . . . . . . $= \;\frac{1}{25}\ln(1-2x) + \frac{1}{5}(1-2x)^{-1} -\frac{1}{25}\ln(3-x) + C$

. . . . . . . . $= \;\frac{1}{5}\!\cdot\!\frac{1}{1-2x} + \frac{1}{25}\bigg[\ln(1-2x) - \ln(3-x)\bigg] + C$

. . . . . . . . $= \;\frac{1}{5(1-2x)} + \frac{1}{25}\ln\left[\frac{1-2x}{3-x}\right] + C$