MHB Partial Fraction Decomposition

[shamieh]

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

$$\int \frac{3t - 2}{t + 1} dt$$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... http://www.slader.com/textbook/9780538497909-stewart-calculus-early-transcendentals-7th-edition/492/exercises/8/#

 

[shamieh]

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]

Also how are they getting rid of the $$(t + 1)$$ in the numerator and just saying $$3 - \frac{5}{(t + 1)}$$ is this magic??

 

[MarkFL]

I would write:

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

 

[soroban]

Hello, shamieh!

You can always use Long Division.

. . \begin{array}{ccccc} &amp;&amp;&amp;&amp; 3 \\<br /> &amp;&amp; --&amp;--&amp;-- \\<br /> t+1 &amp; ) &amp; 3t &amp; - &amp; 2 \\<br /> &amp;&amp; 3t &amp; + &amp; 3 \\<br /> &amp;&amp; -- &amp; -- &amp; -- \\<br /> &amp;&amp;&amp; - &amp; 5 \end{array}\text{Therefore: }\:\frac{3t-2}{t+1} \;=\;3 - \frac{5}{t+1}