Question

\[\int\limits_{1}^{4} (x^3+2x^2+16x+16)/(x^4+16x^2)*dx\]

Answer 1

Solve, is there a semi-easy way to do this - without the PF that is.

Answer 2

maybe a trick to write the numerator as the derivative of the denominator? would still have to split it up i think

Answer 3

fist try to factorise .. that might give clue

Answer 4

just thinking out loud but maybe \[\frac{x^3+16x}{x^4+16x^2}+\frac{2x^2+16}{x^4+16x^2}\] would that make it easier?

Answer 5

\[x(x ^{2}+16) +2x ^{2}+16/x ^{2}(x ^{2}+16)\]

Answer 6

no not really, but i bet we could break it up some way

Answer 7

Does PF= partial fractions?

Answer 8

I broke up it up into 4 different terms, and then began finding the PF decomposition of everyone of the terms.

Answer 9

Yes.

Answer 10

It's under that chapter in my book aswell. (Partial fractions that is)

Answer 11

let's ask wolf ... we might construct solution.

Answer 12

must be an easy trick since it is
\[\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^2+16}\]

Answer 13

if we walk in reverse direction

Answer 14

easiest possible numerators, so probably some easy algebra will get it

Answer 15

I got stuck on trying to decompose the last term :'(

Answer 16

http://www.wolframalpha.com/input/?i=integrate+%28x%5E3%2B2x%5E2%2B16x%2B16%29%2F%28+x%5E4%2B16x%5E2%29
looks like satellite73 was right after all.

Answer 17

or just grind it out
\[\frac{x^3+2x^2+16x+16}{x^4+16x^2}=\frac{a}{x}+\frac{b}{x^2}+\frac{cx+d}{x^2+16}\]
\[x^3+2x^2+16x+16=ax(x^2+16)+b(x^2+16)+(cx+d)x^2\] put x = 0 get \(b=1\) right off the bat

Answer 18

if we were snappy with complex numbers we could put x = 4i, but that is probaby too much

Answer 19

but we have b = 1 so \(bx^2+dx^2=x^2+dx^2=2x^2\) now we know d = 1 as well

Answer 20

and \(16x=16ax\) tells us \(a=1) also

Answer 21

\(a=1\)

Answer 22

Thanks alot! This is probably the most messy integral I've tried solving.

Answer 23

And to go one direction - only to realize your way does not work in the last step - kills :(

Answer 24

yeah these are a pain