Help with integration problem. Need to integrate 1 / ( 1 + x^(3/2)) dx Is there a way to do this without using partial fractions?

Question

Help with integration problem.   Need to integrate 1 / ( 1 + x^(3/2)) dx      Is there a way to do this without using partial fractions?

anonymous · Answer

$$\int\limits_{}^{} 1/(1 + \sqrt[3]{x})  dx$$

amistre64 · Answer

maybe z^2 = x^(1/3)  ??  and see if that gets you a good set up, might not tho

anonymous · Answer

hmm, let me try...

amistre64 · Answer

otherwise you might try to see if wolfram has a good step down of it

amistre64 · Answer

http://www.wolframalpha.com/input/?i=integrate+1+%2F+%28+1+%2B+x^%283%2F2%29%29+dx

anonymous · Answer

tried  a bit but am getting lost in circles.   Wolfram uses partial fractions...let me look at your link...

amistre64 · Answer

if we let:

z^6 = x

6z^5 dz = dx

$$\int\frac{6z^4}{1+z^2}dz$$
and integrate by parts?

anonymous · Answer

aye, that might work.

amistre64 · Answer

it might, but then it might not :)

amistre64 · Answer

you could then take out the constant 6; divide out the rest and do a sum of the dividends

amistre64 · Answer

and that should be a 6z^5 .... typoed it

anonymous · Answer

How did you get z^6?   Doesnt z = x^3/2   turn into z^2/3 = x ?

amistre64 · Answer

$$z^6=x;\ z^{6/3}=z^2$$

$$6\left(\frac{z^4}{4}\frac{z^2}{2}+\frac{1}{2}\int\frac{2z}{z^2+1}\right)$$
$$6\left(\frac{z^4}{4}\frac{z^2}{2}+\frac{tan^{-1}(z)}{2}\right)$$

amistre64 · Answer

$$6\left(\frac{z^4}{4}-\frac{z^2}{2}+\frac{tan^{-1}(z)}{2}\right)$$

amistre64 · Answer

you simply change the variable to z^6 = x to accomodate for the rest

amistre64 · Answer

ive seen it in the books before, but i havent had much practice at it

amistre64 · Answer

and it looks prime for this set up tho

amistre64 · Answer

z^3 - z + z/(z^2+1)
             ------------------
z^2 +1 | z^5
                     -z^3
                              +z

amistre64 · Answer

i see a fauxpaux in the integral i did  ..... urgh

anonymous · Answer

this one is a tough one...making my head hurt! :)

amistre64 · Answer

$$6\left(\frac{z^4}{4}-\frac{z^2}{2}+\frac{1}{2}\int\frac{2z}{z^2+1}\right)$$
$$6\left(\frac{z^4}{4}-\frac{z^2}{2}+\frac{ln|2x+1|}{2}\right)$$

amistre64 · Answer

and of course 2x menas x^2; typing and mathing just dont mix

amistre64 · Answer

im sure it doesnt work in the end, but its a good line to try to figure out :)

anonymous · Answer

I'm a little lost a step here, let me study this for a bit.  Mind watching this thread when its  updated by me?

amistre64 · Answer

if im around or maybe tomorrow