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)]

Answer: 1/[5*(1-2x)] + (1/25)*Ln[(2x-1)/(x-3)]

Printable View

- Mar 17th 2009, 12:03 AMgnameniIntegration 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)]

Answer: 1/[5*(1-2x)] + (1/25)*Ln[(2x-1)/(x-3)] - Mar 17th 2009, 12:21 AMADARSH
I would say you should prefer intigration by parts

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

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

Now $\displaystyle \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

$\displaystyle \frac{1}{-4}[\int{ \frac{A}{(x-1/2)^2}} +\int{ \frac{B}{x-3}}] $ - Mar 17th 2009, 01:14 AMgnameni
I do not think that

dx/(1-2x)^2 equals -1/4(dx/(x-1/2)^2) - Mar 17th 2009, 01:20 AMADARSH
- Mar 17th 2009, 01:30 AMgnameni
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 AMADARSH
If you are trying to say that

$\displaystyle

(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)?

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

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

Put (x-1/2) = t

dx =dt

Hence integral becomes

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

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

doing the same for second integral

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

Let's take it from the top . . .

Quote:

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

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

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

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

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

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

Let $\displaystyle x = 0\!:\quad 1 \;=\;A(3) + B(3) + C(1) \quad\Rightarrow\quad 3A + 3B + C \;=\;1$

. . Substitute [1] and [2]: .$\displaystyle 3A + \tfrac{6}{5} + \tfrac{1}{25} \:=\:1 \quad\Rightarrow\quad\boxed{ A = \text{-}\tfrac{2}{25}}$

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

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

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

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

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