6. INT (2X^2 + 1)/[(X^6 + 2X^4 + 2X^2 + 1)]^(1/2)dx????

7. INT [(2X^6 + 1)]/(X^6)[(1 + X^2)] dx????

8. INT [(2x^2 - 5)]/[(x^4 - 5x^2 + 6)]dx ?????

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- Mar 29th 2007, 09:20 AMberkanatciplease trying to solve
6. INT (2X^2 + 1)/[(X^6 + 2X^4 + 2X^2 + 1)]^(1/2)dx????

7. INT [(2X^6 + 1)]/(X^6)[(1 + X^2)] dx????

8. INT [(2x^2 - 5)]/[(x^4 - 5x^2 + 6)]dx ????? - Mar 29th 2007, 10:00 AMKrizalid
- Mar 29th 2007, 10:05 AMberkanatci
you are right but my exam ll be very soon and I am afraid :(

this weekend ,my friend and I ll try to understand your solving problems:)

bec of not enough time,I cant solve questions.

please help me - Mar 29th 2007, 10:10 AMKrizalid
Create a post which says something about your exam, pick up all the integrals you can't solve and all we're gonna help ya.

- Mar 29th 2007, 10:19 AMberkanatci
ok

thank u for everything... this weekend ,I ll study a lot.I promise u :)

monday or tuesday,I ll send a new mail which I cant solve new problems:) - Mar 29th 2007, 12:38 PMAfterShock
As you have written now, you don't even have an integral.

Careful with parentheses:

int([(2x^6 + 1)/(x^6)]*(1 + x^2)dx)

Is that what you're trying to compute?

If so:

Expand (x^2 + 1)*(2x^6 + 1) = 2x^8 + 2x^6 + x^2 + 1

And thus, (2x^8 + 2x^6 + x^2 + 1)/x^6

Simplify: 2x^2 + 1/x^4 + 1/x^6 + 2

Integrate:**(2x^3)/3 - 1/(3x^3) - 1/(5x^5) + 2x + C** - Mar 29th 2007, 01:48 PMecMathGeek
I think he's actually trying to integrate:

INT (2x^6 + 1)/[(x^6)(1 + x^2)] dx ... the x^6 and 1 + x^2 both being on the denominator.

Most of the problems I have helped him on so far are actually very long and complicated. He has yet to integrate something as simple as some product of polynomials divided by x^6. So you are very correct in that he needs to be careful with parenthesis. - Mar 29th 2007, 10:28 PMberkanatci
[INT (2x^6 + 1)/[(x^6)(1 + x^2)] dx ... this is correct one.

can u solve it - Mar 30th 2007, 06:37 AMecMathGeek
7. INT (2X^6 + 1)/[(x^6)(1 + x^2)] dx

This one is pretty interesting. I won't go through the entire process of solving it, but I will set up the final integration:

First, add and subtract 1 from the numerator

INT (2x^6 + 2 - 1)/[(x^6)(1 + x^2)] dx

INT 2(x^6 + 1)/[(x^6)(1 + x^2)] dx - INT 1/[(x^6)(1 + x^2)] dx

This numerator is a sum of cubes, which can be factored:

2*INT [(x^2 + 1)(x^4 - x^2 + 1)]/[(x^6)(1 + x^2)] dx

2*INT (x^4 - x^2 + 1)/x^6 dx

2*INT (1/x^2 - 1/x^4 + 1/x^6) dx

Add and subtract x^2 from this integration:

INT (1 + x^2 - x^2)/[(x^6)(1 + x^2)] dx

INT (1 + x^2)/[(x^6)(1 + x^2)] dx - INT x^2/[(x^6)(1 + x^2)] dx

INT 1/x^6 - INT 1/[(x^4)(1 + x^2)] dx

Add and subtract x^2 from this integration:

INT (1 + x^2 - x^2)/[(x^4)(1 + x^2)] dx

INT (1 + x^2)/[(x^4)(1 + x^2)] dx - INT x^2/[(x^4)(1 + x^2)] dx

INT 1/x^4 dx - INT 1/[(x^2)(1 + x^2)] dx

Add and subtract x^2 from this integration:

INT (1 + x^2 - x^2)/[(x^2)(1 + x^2)] dx

INT (1 + x^2)/[(x^2)(1 + x^2)] dx - INT x^2/[(x^2)(1 + x^2)] dx

INT 1/x^2 dx - INT 1/(1 + x^2) dx

Combine the entire integration into one (be careful with negative signs) and integrate. It will be much easier to do at this point.

- Mar 30th 2007, 06:55 AMKrizalid