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Partial Functions Integration
- Thread starter hieurock
- Start date
D
Deleted member 4993
Guest
They have used the method of partial fractions (not functions).What is done to be able to get from the first step to the second step in the attached photo? I am not sure how exactly it was derived. Thanks in advance.
Assume:
\(\displaystyle \frac{1}{x * (3x + 1)} \ = \ \frac{A}{x} + \frac{B}{3x+1}\) .......................then
\(\displaystyle \frac{1}{x * (3x + 1)} \ = \ \frac{3A*x + A + B*x}{x * (3x +1)} \ \)
\(\displaystyle \frac{1}{x * (3x + 1)} \ = \frac{x * (3A + B) + A}{x * (3x +1)}\).......................then
equating components of 'x' in The numerator and denominator of LHS and RHS, we get
3A + B = 0 ......... and
A = 1
Solve for 'A' and 'B' and use in
\(\displaystyle \frac{A}{x} + \frac{B}{3x+1}\)
If you are still stuck - write back showing your work.
[MATH]\frac{4}{x(3x+1)}= \frac{?}{x} + \frac{?}{3x+1}[/MATH]To find what number goes over the x, put your finger over the x in the expression on the left hand side. Put in 0 for x: answer is [MATH]\frac{4}{3\times 0 + 1}=4[/MATH]To find what number goes over the (3x+1), put your finger over the (3x+1) in the expression on the left hand side. Put in [MATH]-\frac{1}{3}[/MATH] for x:
answer is [MATH]\frac{4}{-\frac{1}{3}}=-12[/MATH]
(Note: the number you are substituting in for x is the number which makes the covered expression=0.
This method works when you have different linear terms on the bottom e.g. (2x+1)(x)(x-2) etc...)
However you should look up and study the general methods for partial fractions.
answer is [MATH]\frac{4}{-\frac{1}{3}}=-12[/MATH]
(Note: the number you are substituting in for x is the number which makes the covered expression=0.
This method works when you have different linear terms on the bottom e.g. (2x+1)(x)(x-2) etc...)
However you should look up and study the general methods for partial fractions.
Hey guys. I have a question here.
Why [MATH]y[/MATH] is a function of [MATH]1/x[/MATH], ie [MATH]f(\frac{1}{x})[/MATH], instead of [MATH]f(x)[/MATH]?
I am not comfortable of this function. Can I just transform it into [MATH]f(u)[/MATH], for example? If yes, how would I do that?
Why [MATH]y[/MATH] is a function of [MATH]1/x[/MATH], ie [MATH]f(\frac{1}{x})[/MATH], instead of [MATH]f(x)[/MATH]?
I am not comfortable of this function. Can I just transform it into [MATH]f(u)[/MATH], for example? If yes, how would I do that?
D
Deleted member 4993
Guest
Remember, OP was NOT about INTEGRATION - but about partial fraction.Hey guys. I have a question here.
Why [MATH]y[/MATH] is a function of [MATH]1/x[/MATH], ie [MATH]f(\frac{1}{x})[/MATH], instead of [MATH]f(x)[/MATH]?
I am not comfortable of this function. Can I just transform it into [MATH]f(u)[/MATH], for example? If yes, how would I do that?
Sure you can transform it - but that would not answer OP directly.
OP has already got what he needs from the second postRemember, OP was NOT about INTEGRATION - but about partial fraction.
Sure you can transform it - but that would not answer OP directly.
D
Deleted member 4993
Guest
I was responding to your "comfort" of conversion to "u" (=1/x) - instead of working with "x".OP has already got what he needs from the second post