What is the integral of (2400x)sqrt(3+2x) ?

The integral of (2400x)sqrt(3+2x) is 600(2/5 (2x+3)^{5/2}-2(2x+3)^{3/2})+C

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integral of (2400x)sqrt(3+2x)

\int (2400 x) 3 + 2 x d x

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

Solution steps

\int 2400 x 3 + 2 x d x

Take the constant out:

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

= 2400 \cdot \int x 3 + 2 x d x

Apply u-substitution

: \int \frac{u ( u - 3 )}{4} d u

= 2400 \cdot \int \frac{u ( u - 3 )}{4} d u

Take the constant out:

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

= 2400 \cdot \frac{1}{4} \cdot \int u (u - 3) d u

Expand

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

= 2400 \cdot \frac{1}{4} \cdot \int u^{\frac{3}{2}} - 3 u d u

Apply the Sum Rule:

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

= 2400 \cdot \frac{1}{4} (\int u^{\frac{3}{2}} d u - \int 3 u d u)

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

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

= 2400 \cdot \frac{1}{4} (\frac{2}{5} u^{\frac{5}{2}} - 2 u^{\frac{3}{2}})

Substitute back

u = 3 + 2 x

= 2400 \cdot \frac{1}{4} (\frac{2}{5} (3 + 2 x)^{\frac{5}{2}} - 2 (3 + 2 x)^{\frac{3}{2}})

Simplify

2400 \cdot \frac{1}{4} (\frac{2}{5} (3 + 2 x)^{\frac{5}{2}} - 2 (3 + 2 x)^{\frac{3}{2}}) : 600 (\frac{2}{5} (2 x + 3)^{\frac{5}{2}} - 2 (2 x + 3)^{\frac{3}{2}})

= 600 (\frac{2}{5} (2 x + 3)^{\frac{5}{2}} - 2 (2 x + 3)^{\frac{3}{2}})

Add a constant to the solution

= 600 (\frac{2}{5} (2 x + 3)^{\frac{5}{2}} - 2 (2 x + 3)^{\frac{3}{2}}) + C

\frac{d}{d x} (4 + 2 x^{\frac{3}{2}})

\frac{d}{d x} (ln ((x + 3)^{2}))

\int \frac{x ^{3}}{x ^{2} - 36} d x

\frac{\partial}{\partial y} (2 x y)

What is the integral of (2400x)sqrt(3+2x) ?
The integral of (2400x)sqrt(3+2x) is 600(2/5 (2x+3)^{5/2}-2(2x+3)^{3/2})+C