pLEASE EXPLAIN SERIOUSLY HOW TO INTEGRATE A DILOGARITHMIC FUNCTION. BE SERIOUS THIS TIME.?

Question

TAKE ANY DILOGARITHMIC FUNCTION AS EXAMPLE AND INTEGRATE IT. SHOW ALL NECESSARY STEPS USED IN THE INTEGRATION

YOU MIGHT CONSIDER QUESTION CATEGORY AS INVALID BUT YOU ARE ADVISED TO IGNORE IT.

IF YOU ARE UNABLE TO TAKE ANY EXAMPLE THEN PLEASE INTEGRATE UNDERSAID FUNCTION SHOWING ALL NECESSARY STEPS.

log[e, X] / (1 + X^0.33333)

BEFORE ANSWERING PLEASE READ THE NOTES BELOW.

Notes :-
1) log[e, X] MEANS LOGARITHM OF 'X' WITH BASE 'e'.

2) X^0.33333 MEANS CUBE ROOT OF X.

3) DON'T GIVE THIS WEBSITE AS ANSWER www.wikipedia.org/dilogarithmic
AS WIKIPEDIA IS MEANINGLESS WEBSITE HERE. IT IS NOTHING BUT THE WASTAGE OF TIME.

4) I HAVE TRIED GOOGLE SEARCH ALSO BUT TO MY UTTER SURPRISE IT GAVE NO RESULT.

Answer 1

Int log (1+x^1*/3)dx Integrate by parts
= xlog(1+x^1/3) - Int x* 1/(1+x^1/3)* 1/3 *x^-2/3
Now we have the Int x^1/3/(x^1/3+1) dx
If we call x^1/3 = z 1/3*x^-2/3 dx = dz and dx = 3z^2dz so the integral is
Int 3z^3/(1+z) dz = 3 Int (z^2-z+1 -1/(z+1))dz =
3(z^3/3-z^2/2+z- logI 1+zI) = (z^3 -3/2 z^2 +3z-3 logI1+zI
So we get
xlog(1+x^1/3) -1/3(x-3/2 x^2/3 +3x^1/3 -logI1+x^1/3I) between e and X=
Xlog(1+X^1/3) -1/3(X-3/2X^2/3+3X^1/3 -logI1+x^1/3I-
elog(1+e^1/3)+1/3(e-3/2e^2/3 +3e^1/3 -logI1+e^1/3I)

Answer 2

Go to hand book of mathematics and physics or chemistry. It should be in your college library(reference section) or your physics, chemistry and or mathematics professor should have it.

About 43 years a go. I have seen the answer to it, in the hand book of mathematics and chemistry for sure.

log to the base e is understood when written as ln x, so look for (ln x)/{1+(cub root of x)}. Use the laws of integration and laws of logarithm, to break it up in to terms and solve this.

La. Plaz transform may help also.

PS:
Ln is base 10 (Math convetion "L" capital).
ln is base e ( Math covention "l" small).
All others are Log(a, x).

May want to try other sites again. Using correct convention.