Integration problem help... partial fraction...

**sgcb** · January 30th 2010, 11:41 AM

I can't seem to figure out this partial fractions problem:

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

My work so far has involved factoring out a 3x in the denominator so the problem then looks like $\int \frac {1}{3x(1-x)}$ , and then multiplying through using the cover up rule, but so far that method hasn't resulted in the answer:

$(\frac {1}{3})\ln \frac {x}{(1-x)} + c$

**pickslides** · January 30th 2010, 11:44 AM

Did you find A and B where $\frac {1}{3x(1-x)} = \frac {A}{3x} +\frac {B}{1-x}$ ?

**adkinsjr** · January 30th 2010, 11:49 AM

Originally Posted by sgcb

I can't seem to figure out this partial fractions problem:

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

My work so far has involved factoring out a 3x in the denominator so the problem then looks like $\int \frac {1}{3x(1-x)}$ , and then multiplying through using the cover up rule, but so far that method hasn't resulted in the answer:

$(\frac {1}{3})\ln \frac {x}{(1-x)} + c$

I'm not sure what rule your referring to. I would just decompose the function.

$\int\frac{1}{3x(1-x)}dx=\int\left(\frac{A}{3x}+\frac{B}{1-x}\right)dx$

Are you familiar with this method?

$A(1-x)+B(3x)=1$

$x(3B-A)+A=1$

So now you solve the system of equations:
--------------------------

$3B-A=0$

$A=1$
--------------------------

Substitute those values for A and B into the integral, and it's pretty easy from there.

**skeeter** · January 30th 2010, 11:58 AM

Originally Posted by sgcb

I can't seem to figure out this partial fractions problem:

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

My work so far has involved factoring out a 3x in the denominator so the problem then looks like $\int \frac {1}{3x(1-x)}$ , and then multiplying through using the cover up rule, but so far that method hasn't resulted in the answer:

$(\frac {1}{3})\ln \frac {x}{(1-x)} + c$

note that $\frac{1}{3x(1-x)} = \frac{1}{3} \cdot \frac{1}{x(1-x)}$

and ...

$\frac{1}{x(1-x)} = \frac{1}{x} + \frac{1}{1-x} $

**sgcb** · January 30th 2010, 12:03 PM

Originally Posted by adkinsjr

I'm not sure what rule your referring to. I would just decompose the function.

yes both your method and my method output the values A=1 and B=(1/3). But plugging them back into the integral does not yield the correct answer, unless I'm missing a simplification step....

My answer:
$\ln 3x + (1/3)\ln 1-x + c$

Book:

$ (\frac {1}{3})\ln \frac {x}{(1-x)} + c $

**pickslides** · January 30th 2010, 12:07 PM

Recall the fact that

$\ln(a)- \ln(b) = \ln\left(\frac{a}{b}\right)$

**sgcb** · January 30th 2010, 12:13 PM

Originally Posted by pickslides

Recall the fact that

$\ln(a)- \ln(b) = \ln\left(\frac{a}{b}\right)$

I've considered that property, but each time I work out the problem, the natural logs come out as being added and not subtracted.

**pickslides** · January 30th 2010, 12:18 PM

$ \frac{1}{x} + \frac{1}{1-x} =\frac{1}{x} + \frac{1}{-(x+1)}= \frac{1}{x} - \frac{1}{x+1} $

now apply the log law.

**sgcb** · January 30th 2010, 12:21 PM

Originally Posted by skeeter

note that $\frac{1}{3x(1-x)} = \frac{1}{3} \cdot \frac{1}{x(1-x)}$

and ...

$\frac{1}{x(1-x)} = \frac{1}{x} + \frac{1}{1-x} $

when the 1/3 is factored out I get:

$1 = A(1-x) + B(x)$
$= x(B-A) + A$

then
$B - A = 0$
so
$B=A=1$

or am I doing it wrong?

**e^(i*pi)** · January 30th 2010, 12:24 PM

Originally Posted by sgcb

I've considered that property, but each time I work out the problem, the natural logs come out as being added and not subtracted.

I suggest skeeter's method as you don't need to solve partial fractions:

$ \int \frac {1}{3x(1-x)} = \int \frac{1}{3} + \int \frac{1}{x} + \int \frac{1}{1-x} $

==============================================

I've substituted the values of A and B you found into the expression $\int\left(\frac{1}{3x}+\frac{1}{3-3x}\right)dx $

Note that you integrate that second term you will need to divide by -3 because of the coefficient on the x

$\frac{1}{3}\ln(3x) - \frac{1}{3}\ln(3-3x) + C$ where k is a constant

$\frac{1}{3} \ln \left(\frac{3x}{3(1-x)}\right) +C = \frac{1}{3}\ln \left(\frac{x}{1-x}\right) + C$

**skeeter** · January 30th 2010, 12:24 PM

$\int \frac{1}{x} + \frac{1}{1-x} \, dx = \int \frac{1}{x} - \frac{-1}{1-x} \, dx = \ln|x| - \ln|1-x| + C = \ln\left|\frac{x}{1-x} \right| + C$

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