What is the integral of 2+sqrt(6x) ?

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

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integral of 2+sqrt(6x)

\int 2 + 6 x d x

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

Solution steps

\int 2 + 6 x d x

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \int 2 + 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 2 d x + \int 6 x d x

\int 2 d x = 2 x

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

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

Add a constant to the solution

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

n = 1 \sum \infty x^{n + 1}

\frac{\partial}{\partial x} (x \cdot e^{x \cdot y})

\int \frac{5}{4 - ( 2 x ) ^{2}} d x

\int_{2}^{3} \frac{1}{x - 3} d x

y = 3 x^{2} - 4

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