What is the integral of sqrt(x)+1/(6sqrt(x)) ?

The integral of sqrt(x)+1/(6sqrt(x)) is 2/3 x^{3/2}+1/3 sqrt(x)+C

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integral of sqrt(x)+1/(6sqrt(x))

\int x + \frac{1}{6 x} d x

\frac{2}{3} x^{\frac{3}{2}} + \frac{1}{3} x + C

Solution steps

\int x + \frac{1}{6 x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x d x + \int \frac{1}{6 x} d x

\int x d x = \frac{2}{3} x^{\frac{3}{2}}

\int \frac{1}{6 x} d x = \frac{1}{3} x

= \frac{2}{3} x^{\frac{3}{2}} + \frac{1}{3} x

Add a constant to the solution

= \frac{2}{3} x^{\frac{3}{2}} + \frac{1}{3} x + C

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What is the integral of sqrt(x)+1/(6sqrt(x)) ?
The integral of sqrt(x)+1/(6sqrt(x)) is 2/3 x^{3/2}+1/3 sqrt(x)+C