Question

Integral of (5x^2)/((x^2-4)(3x^2+8)). How do you work it with the difference of squares in there?

Answer 1

use equation editor...

\[\int\limits \frac{ 5x^2 }{ \left( x^2-4 \right) \left( 3x^2+8 \right) }\,dx\]like this?

Answer 2

you have to use partial fraction decomposition...

Answer 3

you can factor a difference of squares
a^2-b^2=(a-b)(a+b)

Answer 4

i would hope that @kreimer20 knows that... this is calculus after all!

Answer 5

i know its calculus

Answer 6

i know

Answer 7

he just asked how you do it with a difference of squares
that particular difference of squares can be factored as a linear * linear

Answer 8

when writing as partial fractions
you would do A/linear +B/the other linear +(Cx+D)/the quadratic you have there

Answer 9

yes, and as student studying calculus, she or he should know how to factor a difference of squares

Answer 10

some people forget algebra
just reminding him just in case

Answer 11

i have seen students do some crazy algebra things in calculus

Answer 12

like as in crazy wrong

Answer 13

and unfortunately, some never learned.

me, too! it's so funny (sic) that people struggle with calculus because they don't know their algebra or trig!

Answer 14

i'm sure there was a couple of things i forgot when i was taking calculus (i think :p)

Answer 15

I always found Difference of Cubes formula hard to remember :) lol

Answer 16

also, see Paul's online math notes (here's a link).

The site does a great job of working through the partial fraction decomposition. The attachments I sent previously are also a great resource!

http://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx

Answer 17

at one point i was like i'm don't going to learn that formula
i can find one zero for a difference or sum of cubes then use that to find the other factor by using division

Answer 18

I know how to factor the difference of squares... Just forgot that I could do that to this problem. My professor has made it a little confusing for me...

Answer 19

sum and difference of cubes is easy if you remember to repeat with out cubes, and then perfect square trinomial with different sign and no doubling of middle term...

\[\left( a^3+b^3 \right)=\left( a+b \right)\left( a^2-ab+b^2 \right)\]
\[\left( a^3-b^3 \right)=\left( a-b \right)\left( a^2+ab+b^2 \right)\]

Answer 20

yeah at one point
i know those by heart now seen them for like 8000 years

Answer 21

so @kreimer20 i guess you can go further on this problem now? let me or us know if you need further help.

Answer 22

so, @kreimer20 are you good? hold on to those pdfs and the link to Paul's online math notes.