Find the integral of x/ (x^4+x^2+1) from 2 to 3.

Question

I was going to integrate using the method for partial fractions but I can't factor x^4+x^2+1.

Answer 1

integrate x/(x^4+x^2+1)
| = integrate symbol
u = x^2
du = 2x dx
1/2 du = x dx

1/2 | du/(u^2 + u + 1)

Complete the square
u^2 + u = -1
u^2 + u + 1/4 = -1 + 1/4
(u + 1/2)^2 + 3/4

1/2 | du/((u + 1/2)^2 + 3/4)

w = u + 1/2
dw = du
1/2 | dw/(w^2 + 3/4)
1/2 | dw/(3/4 + w^2)

And, since,
| dx/(a^2 + x^2)= 1/a arctan x/a + C

the integration can easily be done now.

Can you take it from here?

Answer 2

Okay, so I ended up with the expression:

1/2[4/3arctan(4(x^2+1/2)/3)

Can you confirm this is correct before I use the fundamental theorem of calculus to determine the definite integral please.

Answer 3

1/2[2/(sqrt3)arctan(2(x^2+1/2)/(sqrt3)

I distributed the 1/2, but backed it out for my above answer...I don't think I made any errors backing it out.

Here is the answer, with the 1/2 distributed. I know this answer is right because I double check it with an online integration calculator.

1/(sqrt3)arctan (2x^2 + 1)/(sqrt3)

I didn't rationalize the denominators, because the online answer didn't.

Do you want me to post the final answer for you to check the definite integral?

Answer 4

That would be great if you could!
I'm working on getting the correct indefinite as you have above, but when I get it it would be great to have something to check my definite integral with :)

Answer 5

1/(sqrt3)arctan(5/(29(sqrt3))= 0.0572826

Check you values for 1/a and x/a
maybe that's where you got hung up
| dx/(a^2 + x^2)

(3/4 + w^2)
a^2 = 3/4
a = (sqrt3)/2
x^2 = w^2
x = w

So, you should have got,
1/2(1/(sqrt3/2)arctan (w/(sqrt3/2))
which simplified is,
1/2(2/(sqrt3)arctan(2w/(sqrt3))

then I distributed the 1/2
2/(2(sqrt3))arctan(2w/(sqrt3))
1/(sqrt3)arctan(2w/(sqrt3))

then substitute back in for u and x

Good Luck!!