1. ## Simpl- integral

Calculate :

2. Originally Posted by apcalculus
you didn't post anything

Originally Posted by dhiab
Calculate : $\int_0^1 \frac {x^3}{x^6 + 2x^3 + 1}~dx$
it seems like a pain, but i don't see an easier way to approach this at the moment. here's what i'd do

write as $\int_0^1 \frac {x^3}{(x^3 + 1)^2}~dx = \int_0^1 \frac {x^3}{(x + 1)^2(x^2 - x + 1)^2}~dx$

now begin the painful method of partial fractions

3. What? It can be easily factored as $(x^3 + 1)^2$.

4. Originally Posted by Kaitosan
What? It can be easily factored as $(x^3 + 1)^2$.
that has already been noted. are you saying you see an easier way to proceed than partial fractions?

5. Maybe trigonometric substitution would be easier, I don't know.

6. Originally Posted by Kaitosan
Maybe trigonometric substitution would be easier, I don't know.
trig substitution requires one of the forms: $x^2 + a^2$, $x^2 - a^2$, or $a^2 - x^2$ for some nonzero constant $a$. none of those forms show up here

7. Wow. I really do think that's wrong. Trig sub doesn't have that form. It has the form of $u^2 + a^2$ and similar ones in which "u" is an expression that involves a variable like "x".

8. My fault, I misread this completely!

9. I'm confused. What point did I make? lol

I'm not going out to do all the math but here's the bottom line -

In some cases, either partial fractions as well as trig sub can be used. I believe this is one of those cases in which $u = x ^ (3/2)$. Then you solve the eqution u=a(one trig expression) isolating x.

10. Originally Posted by HallsofIvy
No, $(x+1)^2$ is NOT equal to [tex](x+1)^2(x^2- x+ 1). That was Kaitosan's point!
What Jhevon wrote is correct... What you copied is ... a nonsense ?

@ Kaitosan : you need to factorise into (irreducible) polynomials with smaller degree

11. ## Mr F's case study in Diplomacy and Tact

Originally Posted by HallsofIvy
No, $(x+1)^2$ is NOT equal to [tex](x+1)^2(x^2- x+ 1). That was Kaitosan's point!
Since it's a little known fact that moo is studying Diplomacy and Tact 101 and failing badly (the former is the little known fact, by the way), I will jump in here using HallsofIvy's post as a case study:

*Ahem* I think Jhevon said that $(x^3 + 1)^2 = (x + 1)^2 (x^2 - x + 1)^2$, which is a correct identity. His point in saying this was to get the linear and quadratic factors required for the denominators of the partial fraction decomposition.

Kaitosan's point (and just because you don't think you made one doesn't mean you didn't ) was that a trignometric substitution can be made in a wider range of forms than Jhevon initially listed. Note: For some of these forms a hyperbolic substitution probably provides an easier path.

I'm thinking of using this example in my next book: Diplomacy and Tact for Bovines. (There will be a French edition). (And there will be an Appendix on latex tags .... )

You may wish to view my reply in this thread for a further example: http://www.mathhelpforum.com/math-he...tml#post331889

12. Haha !

No, in this case, the problem is that I didn't understand what HallsofIvy wanted to say, as I misread "Kaitosan's point" and with the typos, it's not easier...

*Ahem* I think Jhevon said that $(x^3 + 1)^2 = (x + 1)^2 (x^2 - x + 1)^{\color{red}2}$, which is a correct identity.
Better this way

13. Originally Posted by dhiab
Calculate :
put $\frac{3x^2}{(x^3+1)^2} \ dx = dv, \ \frac{1}{3}x=u.$ then $v=\frac{-1}{x^3+1}$ and $du=\frac{1}{3}dx.$ applying integration by parts will give you $\int_0^1 \frac{x^3}{(x^3 + 1)^2} \ dx = \frac{-1}{6} + \frac{1}{3}\int_0^1 \frac{dx}{x^3 + 1}.$ i guess you can finish the job now.

the final answer, according to my calculation, is $\frac{\pi \sqrt{3}}{27} + \frac{1}{9} \ln 2 - \frac{1}{6}.$

14. If the upper limit of the integral is infinity , it will be very well

15. Originally Posted by Moo
What Jhevon wrote is correct... What you copied is ... a nonsense ?

@ Kaitosan : you need to factorise into (irreducible) polynomials with smaller degree
That's only if I want to use partial fraction integration!!!! I can use trig sub as well if I want.

Edit: Aah, I see. Trig sub is only valid if you can solve the integral of $(tanx)^(2/3)$ lol