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}{d x} ((3 x)^{l n (3 x)})

\frac{d r}{d t} = - 2 t r, r (0) = 2

\int sin^{- \frac{3}{2}} (x) cos^{3} (x) d x

x \to - 1 lim (2 x^{2} - 5 x + 3)

\int \frac{1}{4 + t} d t

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