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Answer 1 · 2017-01-10T13:20:43

# I=2x^(1/2)-3x^(1/3)+6x^(1/6)-6ln|x^(1/6)+1|+C.#

Let #I=int1/(x^(1/2)+x^(1/3))dx#

Observe that, the lcm of #2 and 3# is #6#, so, we select the

substitution #x=t^6," giving, "dx=6t^5dt, x^(1/2)=t^3, x^(1/3)=t^2#.

#:. I=int(6t^5)/(t^3+t^2)dt=6intt^3/(t+1)dt=6int(t^3+1-1)/(t+1)dt#

#=6int[(t^3+1)/(t+1)-1/(t+1)]dt=6int[t^2-t+1-1/(t+1)]dt#

#=6{t^3/3-t^2/2+t-ln|t+1|}#

#=2t^3-3t^2+6t-6ln|t+1|,# and as #t=x^(1/6)#,

#:. I=2x^(1/2)-3x^(1/3)+6x^(1/6)-6ln|x^(1/6)+1|+C.#

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