Posted by Rachel on Saturday, January 22, 2011 at 8:11pm.
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?
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.
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?
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 :)
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!!