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)]
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)]
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}}] $
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,try again
Twinkle Twinkle Little Star..
Before solving question you have asked you must have solved these types of questionHow 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
$\displaystyle \int{\frac{dx}{x-3}} = ln|x-3| +c$
Hllo, gnameni!
Let's take it from the top . . .
Partial Fractions . . .$\displaystyle \int \frac{dx}{(1-2x)^2(3-x)}$
$\displaystyle \text{Answer: }\;\frac{1}{5(1-2x)} + \frac{1}{25}\ln\left[\frac{2x-1}{x-3}\right]$
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 $