Calculus Integration: Solving a Tricky Homework Problem

[imataxslave]

Homework Statement

Int x/x^4+x^2+1 dx

Homework Equations

u=x^2 dx=2x

The Attempt at a Solution

I tried u=x^2+x+1 then du=2x+1
1/2 int u+1/u^2 du
I thought I had it because it made a nice u/u^2 and 1/u^2
But no dice says the prof.
He said let u=x^2 that gives me 1/2 int du/u^2+u+1/u but I'm stuck.
Maybe partial fractions?

 

[lanedance]

hey imataxslave, welcome to pf - you should try & use brackets to make it clear what your equations actually are & make it easy for people to read

ie x/x^4+x^2+1 = (1/x^3 + x^2 +1) to me, so you should write it as
x/(x^4+x^2+1)

though guessing at what you have done, you get to
\int \frac{du}{u^2 + u + 1}

now i haven't tried it, but you could try completing the square on the denominator, then substututing for whatever is in the the square part - should hopefully take it to a more familiar form

 

[lanedance]

updated above

 

[imataxslave]

Thanks. Sorry about the mess. This is only my fourth math class. Now that I'm sharing with others I've been getting many requests to quit being so sloppy. I tried to complete the square but the one cancels out and I' back to the original input. I think I remember a trick using fractions, I'll look into it. This at least gives me a direction to go.

 

[imataxslave]

thanks for the idea! Problem solved.
let u =x^4+x^2+1
du= 4x^3+2x dx
Complete the square
(x^2)^2+(2x^2/2)+1/4+1-1/4 dx
=int x/(((x^2+1/2)^2)+(3/4)) dx
let u^2=(x^2+1/2)
du=2xdx
=int 1/2 du/(u^2+3/4)
=1/2 int du/(u^2+(sqrt3/2)^2)
that's the form I need because dx/(x^2+a^2)= (1/a)arctan (x/a) + c
all that's left is plugging it in and changing back substitutions.
I hope this helps others. It's going in my book of half day problems.