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 ?????

2. Originally Posted by berkanatci
INT [(2x^2 - 5)]/[(x^4 - 5x^2 + 6)]dx
You post many integrals every day.

If you really wanna learn, take it easy, it's much better.

Notice that x^4 - 5x² + 6 = (x² - 3)(x² - 2), and (x² - 3) + (x² - 2) = 2x² - 5; then split the integral in two.

3. 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.

4. 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.

5. 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

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

7. Originally Posted by AfterShock
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
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.

8. [INT (2x^6 + 1)/[(x^6)(1 + x^2)] dx ... this is correct one.

can u solve it

9. Originally Posted by berkanatci
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 ?????
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.

10. Originally Posted by berkanatci
[INT (2x^6 + 1)/[(x^6)(1 + x²)] dx ... this is correct one.

can u solve it
Split the integral in two, you'll get one with 2x^6/(x^6(x² + 1)), which is easy to solve, and 1/(x^6(x² + 1)) = (x^4 - x² + 1)/x^6 - 1/(x² + 1) (see why)

Then the second one it's easy to solve and the problem it's done.