How do you integrate #int (sqrtx+1/(2sqrtx))dx#?

1 Answer
Dec 21, 2016

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

Explanation:

Start by simplifying the equation.

#=> int(x^(1/2))dx + int(1/2x^(-1/2))dx#

Use the formula #x^ndx = (x^(n + 1))/(n + 1) + C#

#=> 2/3x^(3/2) + 1/2int(x^(-1/2))dx#

You don't have to add #+C# just yet... It should be alright if you add that once you have integrated all terms.

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

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

Hopefully this helps!