What the anti derivative of #int#(#sqrtx/2#+#2/sqrtx#)?

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Answer 1 · 2017-12-16T09:41:29

# 1/3*x^(3/2)+4*x^(1/2)+C.#

I hope, the Question is to find #int(sqrtx/2+2/sqrtx)dx.#

Recall that, #intx^ndx=x^(n+1)/(n+1)+c, where, n ne-1.#

#:. I=int{1/2*x^(1/2)+2*x^(-1/2)}dx,#

#=1/2*x^(1/2+1)/(1/2+1)+2*x^(-1/2+1)/(-1/2+1),#

#=1/2*x^(3/2)/(3/2)+2*x^(1/2)/(1/2),#

#rArr I=1/3*x^(3/2)+4*x^(1/2)+C.#

Enjoy Maths.!

N.B:- If the Question is to find the anti-derivative of

#int(sqrtx/2+2/sqrtx)dx,# then, we have to re-integrate #I!#