What is the integral of (sqrt(x)+3)^2 ?

The integral of (sqrt(x)+3)^2 is (x^2)/2+4x^{3/2}+9x+C

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integral of (sqrt(x)+3)^2

\int (x + 3)^{2} d x

\frac{x ^{2}}{2} + 4 x^{\frac{3}{2}} + 9 x + C

Solution steps

\int (x + 3)^{2} d x

Expand

(x + 3)^{2} : x + 6 x + 9

= \int x + 6 x + 9 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 6 x d x + \int 9 d x

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

\int 6 x d x = 4 x^{\frac{3}{2}}

\int 9 d x = 9 x

= \frac{x ^{2}}{2} + 4 x^{\frac{3}{2}} + 9 x

Add a constant to the solution

= \frac{x ^{2}}{2} + 4 x^{\frac{3}{2}} + 9 x + C

What is the integral of (sqrt(x)+3)^2 ?
The integral of (sqrt(x)+3)^2 is (x^2)/2+4x^{3/2}+9x+C