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

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

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integral of xsqrt(3x+2)

\int x 3 x + 2 d x

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

Solution steps

\int x 3 x + 2 d x

Apply u-substitution

: \int \frac{u ( u - 2 )}{9} d u

= \int \frac{u ( u - 2 )}{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 - 2) d u

Expand

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

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

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

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

= \frac{1}{9} (\frac{2}{5} u^{\frac{5}{2}} - \frac{4}{3} u^{\frac{3}{2}})

Substitute back

u = 3 x + 2

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{2 x + x y ^{2}}{4 y + y x ^{2}}

\int_{0}^{\frac{1}{2}} 6 arccos (x) d x

x \to \infty lim (5 5^{x})

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

y = (1 + 3 x)^{10}, (0, 1)

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