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

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

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

\int \frac{x}{3 x + 2} d x

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

Solution steps

\int \frac{x}{3 x + 2} d x

Apply u-substitution

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

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

Take the constant out:

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

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

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

\int 2 d u = 2 u

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

Substitute back

u = 3 x + 2

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

Simplify

\frac{2}{9} (\frac{( 3 x + 2 ) ^{3}}{3} - 2 3 x + 2) : \frac{2}{9} (\frac{1}{3} (3 x + 2)^{\frac{3}{2}} - 2 3 x + 2)

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

Add a constant to the solution

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

r \to 0 lim (r^{2} ln (r^{2}))

\int 8 z d z

x \to \infty lim (2 (- 23 e^{- x} + 23 e^{- 1}))

\frac{\partial}{\partial x} (e^{y l n (z)})

\int (x^{2} + x^{- 2}) d x

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