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

The integral of x(sqrt(3x+4)) is 1/9 (2/5 (3x+4)^{5/2}-8/3 (3x+4)^{3/2})+C

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integral of x(sqrt(3x+4))

\int x (3 x + 4) d x

\frac{1}{9} (\frac{2}{5} (3 x + 4)^{\frac{5}{2}} - \frac{8}{3} (3 x + 4)^{\frac{3}{2}}) + C

Solution steps

\int x 3 x + 4 d x

Apply u-substitution

: \int \frac{u ( u - 4 )}{9} d u

= \int \frac{u ( u - 4 )}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int u (u - 4) d u

Expand

u (u - 4) : u^{\frac{3}{2}} - 4 u

= \frac{1}{9} \cdot \int u^{\frac{3}{2}} - 4 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

= \frac{1}{9} (\int u^{\frac{3}{2}} d u - \int 4 u d u)

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

\int 4 u d u = \frac{8}{3} u^{\frac{3}{2}}

= \frac{1}{9} (\frac{2}{5} u^{\frac{5}{2}} - \frac{8}{3} u^{\frac{3}{2}})

Substitute back

u = 3 x + 4

= \frac{1}{9} (\frac{2}{5} (3 x + 4)^{\frac{5}{2}} - \frac{8}{3} (3 x + 4)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{9} (\frac{2}{5} (3 x + 4)^{\frac{5}{2}} - \frac{8}{3} (3 x + 4)^{\frac{3}{2}}) + C

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What is the integral of x(sqrt(3x+4)) ?
The integral of x(sqrt(3x+4)) is 1/9 (2/5 (3x+4)^{5/2}-8/3 (3x+4)^{3/2})+C