What is the integral of (sqrt(3x)+1)^2 ?

The integral of (sqrt(3x)+1)^2 is (3x^2)/2+4/(sqrt(3))x^{3/2}+x+C

integral of (sqrt(3x)+1)^2

\int (3 x + 1)^{2} d x

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

Solution steps

\int (3 x + 1)^{2} d x

Expand

(3 x + 1)^{2} : 3 x + 23 x + 1

= \int 3 x + 23 x + 1 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 3 x d x + \int 23 x d x + \int 1 d x

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

\int 23 x d x = \frac{4}{3} x^{\frac{3}{2}}

\int 1 d x = x

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

Add a constant to the solution

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

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What is the integral of (sqrt(3x)+1)^2 ?
The integral of (sqrt(3x)+1)^2 is (3x^2)/2+4/(sqrt(3))x^{3/2}+x+C