∫(x3m+x2m+xm)(2x2m+3xm+6)1/mdx
note: do without using integration by parts
1357(2x2m+3xm+6)1/m=(1/x)(2x3m+3x2m+6xm)1/m
(x3m+x2m+xm)(2x2m+3xm+6)1/m=(x3m+x2m+xm)(1/x)(2x3m+3x2m+6xm)1/m
=((x3m+x2m+xm)/x)(2x3m+3x2m+6xm)1/m=
(x3m-1+x2m-1+xm-1)(2x3m+3x2m+6xm)1/m
rest you can do it
11multiply and divide by x
it becomes
∫1/x(x3m+x2m+xm)(2x3m+3x2m+6xm)1/mdx
now
take 6 common in the 2nd bracket
∫1/x 61/m(x3m+x2m+xm)(1/3x3m+1/2x2m+xm)1/mdx
now
∫61/m(x3m-1+x2m-1+xm-1)(1/3x3m+1/2x2m+xm)dx
61/m∫...............................
now u can proceed
too lazy to proceed
yime 4 lunch :P
1put\ x^m=t\\ \\ x^mdx=\frac{t^{\frac{1}{m}}dt}{m}\\ \\ I=\int (x^{2m}+x^m+1)(2x^{2m}+3x^m+6)^{\frac{1}{m}}x^m\ dx\\ \\ I=\frac{1}{m}\int (t^2+t+1)(2t^3+3t^2+6t)^{\frac{1}{m}}dt\\ \\ put\ 2t^3+3t^2+6t = u\\ \\ 6(t^2+t+1)dt = du\\ \\ I=\frac{1}{6m}\int u^{\frac{1}{m}}du=\frac{u^(\frac{m+1}{m})}{6(m+1)}+C
put back u in terms of x and you will get it...
1sorry guy's didnt saw that u have posted it while i was typing :)
