Question

Calculus 2 help please. I cannot seem to figure out integration of rational functions by partial fractions. I can solve for A and B but don't know what to do with them afterwards. How do i incorporate them back into the equation that needs to be integrated?

Answer 1

for example \[\int\limits_{}^{}\frac{ 2x+13 }{ (x-6)(x-1) }\]
I know A = 5 and B = -3
what do I do with these numbers?

Answer 2

I'd like to know where you got that A=5 and that B=-3.

The objective here is to split the given integrand into two simpler fractions ("partial fractions") that are relatively easy to integrate.

The integrand of your \[\int\limits\limits_{}^{}\frac{ 2x+13 }{ (x-6)(x-1) }\]

Answer 3

would be split up into these two fractions: A/(x-6) and B(x-1).

If A=5 and B=-3, you'd end up with 5/(x-6) and -3/(x-1).

The latter are quite easy to integrate.

Please demo how you got those values for A and B.

Answer 4

i used the equation \[\frac{ 2x+13 }{ (x-6)(x-1)}=\frac{ A }{(x-6)}+\frac{ B }{ (x-1)}\]
\[\frac{ 2x+13 }{ (x-6)(x-1) }=\frac{ A(x-1)+B(x-6) }{ (x-1)(x-6) }\]
canceld out the bottom
\[(2x+13) = a(x-1)+ b(x-6)\]
then canceld out b to find a first
let x=6
\[(2(6)+13)= a(6-1)=b(6-6) = a(5)+b(0)\]
\[a= 25/5 = 5\]
let x= 1 solve for b
\[(2(1)+13)= a(1-1)=b(1-6)= a(0)+b(-5)\]
\[b = 25/-5= -3\]

Answer 5

At first glance you appear to be doing precisely the right thing. Need further explanation and/or help?

Answer 6

The result of the decomposition is that you can write
\[\int\frac{2x+13}{(x-6)(x-1)}\,\mathrm dx=\int\left(\frac{5}{x-6}-\frac{3}{x-1}\right)\,\mathrm dx\]

Answer 7

... and integrate one term at a time (is what I meant to finish that comment with)

Answer 8

Okay so are the so a and b will replace the numerator because we equaled it to original numerator? My professor only did one problem and forgot to show us what to do with a and b.

Answer 9

Backing up one more step:

Partial fraction decomposition is an attempt to write, in this case, some rational function as a sum of "simpler" rational functions
\[\frac{2x+13}{(x-6)(x-1)}=\frac{A}{x-6}+\frac{B}{x-1}\](Notice how the numerator of each partial fraction is of one degree less than that of the denominator.)

I'd be careful about saying that you're "replacing the numerator", because that's not really the case. You're also "replacing" the denominator, but more accurately, you're not replacing anything. Rather, you're just rewriting the original function in an equivalent form in terms of rational functions whose denominators are factors of the original function's denominator.

So now that you have
\[\int\frac{2x+13}{(x-6)(x-1)}\,\mathrm dx=\int\left(\frac{5}{x-6}-\frac{3}{x-1}\right)\,\mathrm dx\]you can find the integral by computing the remaining simpler integrals. In particular, \(\displaystyle\int\frac{\mathrm dx}{x-k}=\ln|x-k|+C\).