All binomial are A: I=∫xMdx(xN+k)P We do x=th

dx=thM th-1t=dt·th(M+1)-1 I=∫dt·t(h(M+1)-1[tNh+k]P we do hN=2

I=∫dt·t2(M+1)/N-1)[t2+k]p we do interest there, is exponent 2

and when k/A2 will be 1 I=∫dwchw·shqw·ch2pw

Now this two factors remind sh·ch when we do t=Ashw if k>0 (if not we do

t=Achw) I=∫dw·ch[2(M+1)/N-1]w·sh2P+1

Cfr. Wikipedia series of shmchn when m & n>0, nm=2P+1 n=2(M+1)/N-1

B: similarly sinmcosn, when I=∫dt·t[h(M+1)N-1][k-tNh]P t=Asinw iffis

[k+tNh]P then t=Ashw. I=∫dwshqw·ch2P+1w q=2(M+1)/N-1

I=∫dw·sinqw·cos2p+1w n=2p+1 m=2(M+1)/N both enters P can be ½, 3/2, 5/2.. etc but also 2(M+1)/N (M+1)/N can be multiple of ½ like P. Much more than Chebishev!!!

A) HIPERBOLICS k>0 shmwchnw, they are integrated in wikipedia.For products of hyperbolics see at the end of the anex:

I(m,n)= ∫du·shmu·chnu=solution when m & n >0 (both enters)

=shm-1u·chn+1u/(m+n)-(m-1)/(n+1) ∫shm-2u·chnu·du h=2/N Mh+h-1=(2M/N+2/N-N/N=2(M+1)/N-1

I=∫d(shw)sh[2(M+1)N-1]w[sh2w+1]P, if k<0 ;,we would do t=Achw

I=∫dwchwsh[2M+1)/N-1]w·ch2Pw=∫dwsh[2M+1)/M-1]w·ch2P+1w wikipedia solves

I=∫dz·zM(z2-1)P the lists of wikipedia only ask for enter exponents. 2P exponent for ch, and M for sh (or when ∫dz·zM(z2+1)P then M is for sh and 2P for ch

If one exponent is negative we have other formulas:

And the circulars

The condition for being integrable is than n and m were enters.

M &N enters. m and p can be, besides enters, a half of an enter.

(Chebyshev believed to have demonstrated that there was no more cases, than his three enters)