What is the antiderivative of #6sqrtx#?

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Answer 1 · 2016-04-16T21:47:43

#int6sqrt(x)dx = 4x^(3/2)+C#

In general, we have #intx^kdx = x^(k+1)/(k+1)+C# for #k!=-1#. Putting that together with #intkf(x)dx = kintf(x)dx# we have:

#int6sqrt(x)dx = 6intx^(1/2)dx = 6x^(3/2)/(3/2)+C = 4x^(3/2)+C#

Answer 2 · 2016-04-16T21:49:21

#4x^(3/2)+C#

We want to find

#int6sqrtxdx#

Which, since #6# is a multiplicative constant, equals

#=6intsqrtxdx#

Write #sqrtx# using fractional exponents:

#=6intx^(1/2)dx#

Now, integrate using the rule:

#intx^ndx=x^(n+1)/(n+1)+C" "" "," "" "n!=-1#

Giving:

#=6(x^(1/2+1)/(1/2+1))+C=(6x^(3/2))/(3/2)+C=2/3(6x^(3/2))+C#

#=4x^(3/2)+C#