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March 31, 2015

March 31, 2015

Posted by **Rachel** on Saturday, January 22, 2011 at 8:11pm.

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

- Calculus -
**helper**, Saturday, January 22, 2011 at 9:06pmintegrate 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?

- Calculus -
**Rachel**, Saturday, January 22, 2011 at 9:52pmOkay, 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.

- Calculus -
**helper**, Saturday, January 22, 2011 at 10:08pm1/2[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?

- Calculus -
**Rachel**, Saturday, January 22, 2011 at 10:13pmThat 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 :)

- Calculus -
**helper**, Saturday, January 22, 2011 at 10:36pm1/(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!!

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