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Partial Fraction Decomposition

shamieh

Active member
Sep 13, 2013
539
Quick question... I know that if the numerator is greater than the denominator I need to divide out by long division BUT If the numerator is equal to the denominator (the exponent is what I'm talking about to be specific) then, do I need to do anything? Because I'm stuck on this problem

\(\displaystyle \int \frac{3t - 2}{t + 1} dt\)


Some how they are getting like 3(t-5) + 1 or something weird.. I don't understand..What is the first step I should do..

Some how they are changing it... Here it is if you'd like to see it. They aren't doing long division they are doing something else weird... Solution they are getting here
 

shamieh

Active member
Sep 13, 2013
539
I mean now I see what they're doing but I don't see how I'm supposed to just KNOW that 3(t + 1) - 5 is another form of 3t - 2
 

shamieh

Active member
Sep 13, 2013
539
Also how are they getting rid of the \(\displaystyle (t + 1)\) in the numerator and just saying \(\displaystyle 3 - \frac{5}{(t + 1)}\) is this magic??
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would write:

\(\displaystyle \frac{3t-2}{t+1}=\frac{3t+3 - 5}{t+1}=\frac{3(t+1)}{t+1}-\frac{5}{t+1}=3-\frac{5}{t+1}\)
 

soroban

Well-known member
Feb 2, 2012
409
Hello, shamieh!

You can always use Long Division.

. . [tex]\begin{array}{ccccc} &&&& 3 \\
&& --&--&-- \\
t+1 & ) & 3t & - & 2 \\
&& 3t & + & 3 \\
&& -- & -- & -- \\
&&& - & 5 \end{array}[/tex]


[tex]\text{Therefore: }\:\frac{3t-2}{t+1} \;=\;3 - \frac{5}{t+1}[/tex]