What is the integral of (4x(sqrt(3x)-2)) ?

The integral of (4x(sqrt(3x)-2)) is 4(sqrt(3) 2/5 x^{5/2}-x^2)+C

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integral of (4x(sqrt(3x)-2))

\int (4 x (3 x - 2)) d x

4 (3 \frac{2}{5} x^{\frac{5}{2}} - x^{2}) + C

Solution steps

\int 4 x (3 x - 2) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int x (3 x - 2) d x

Expand

x (3 x - 2) : 3 x^{\frac{3}{2}} - 2 x

= 4 \cdot \int 3 x^{\frac{3}{2}} - 2 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

= 4 (\int 3 x^{\frac{3}{2}} d x - \int 2 x d x)

\int 3 x^{\frac{3}{2}} d x = 3 \frac{2}{5} x^{\frac{5}{2}}

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

= 4 (3 \frac{2}{5} x^{\frac{5}{2}} - x^{2})

Add a constant to the solution

= 4 (3 \frac{2}{5} x^{\frac{5}{2}} - x^{2}) + C

\frac{d}{d x} (2 a arccos (2 x))

sin^{4} (6 x^{2})

\int 11 x^{\frac{4}{7}} + 8 x^{- \frac{6}{7}} d x

\int \frac{sec ( θ )}{cos ( θ ) - sec ( θ )} d θ

\frac{d y}{d x} = 1 - y^{2}

What is the integral of (4x(sqrt(3x)-2)) ?
The integral of (4x(sqrt(3x)-2)) is 4(sqrt(3) 2/5 x^{5/2}-x^2)+C