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Answer 1 · 2017-05-12T00:27:27

#=(8sqrtx)/3 - sqrt3/(3x^3) + 5/(12x^2) + (4x^5)/35 +C#

#int(4/3x^(-1/2)+sqrt3x^(-4)-5/(6x^3)+4/(7x^(-4)))dx#

Rewrite with no #x#'s in the denominators:
#=int( 4/3 x^(-1/2)+sqrt3 x^(-4)-5/6 x^(-3)+4/7 x^4)dx#

Integrate using the "opposite of" the differentiation power rule:

#=8/3x^(1/2)-sqrt3/3 x^(-3)+5/12x^(-2)+4/35x^5 +C#

For the sake of better notation, only use positive exponents:
#=(8sqrtx)/3 - sqrt3/(3x^3) + 5/(12x^2) + (4x^5)/35 +C#