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:

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

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

What is the antiderivative of 6sqrtx6sqrtx?