Calculate :
you didn't post anything
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 $\displaystyle \int_0^1 \frac {x^3}{(x^3 + 1)^2}~dx = \int_0^1 \frac {x^3}{(x + 1)^2(x^2 - x + 1)^2}~dx$
now begin the painful method of partial fractions
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 $\displaystyle u = x ^ (3/2)$. Then you solve the eqution u=a(one trig expression) isolating x.
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 $\displaystyle (x^3 + 1)^2 = (x + 1)^2 (x^2 - x + 1)^2$, 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
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...
Better this way*Ahem* I think Jhevon said that $\displaystyle (x^3 + 1)^2 = (x + 1)^2 (x^2 - x + 1)^{\color{red}2}$, which is a correct identity.
put $\displaystyle \frac{3x^2}{(x^3+1)^2} \ dx = dv, \ \frac{1}{3}x=u.$ then $\displaystyle v=\frac{-1}{x^3+1}$ and $\displaystyle du=\frac{1}{3}dx.$ applying integration by parts will give you $\displaystyle \int_0^1 \frac{x^3}{(x^3 + 1)^2} \ dx = \frac{-1}{6} + \frac{1}{3}\int_0^1 \frac{dx}{x^3 + 1}.$ i guess you can finish the job now.
the final answer, according to my calculation, is $\displaystyle \frac{\pi \sqrt{3}}{27} + \frac{1}{9} \ln 2 - \frac{1}{6}.$