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

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Answer 1 · 2016-12-21T00:29:55

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

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!