What is the integral of (x+1)(sqrt(3x+2)) ?

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

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integral of (x+1)(sqrt(3x+2))

\int (x + 1) (3 x + 2) d x

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

Solution steps

\int (x + 1) 3 x + 2 d x

Apply u-substitution

: \int \frac{u ( u + 1 )}{9} d u

= \int \frac{u ( u + 1 )}{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 + 1) d u

Expand

u (u + 1) : u^{\frac{3}{2}} + u

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

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

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

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

Substitute back

u = 3 x + 2

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

f (x) = x^{3} - 12 x

\int (\frac{3}{5}) cos (x) d x

\frac{d}{d x} (\frac{2 x + 1}{x + 2})

\frac{d}{d x} (ln (\frac{7 + e ^{x}}{7 - e ^{x}}))

\int_{0}^{5} \frac{1}{x ^{\frac{13}{12}}} d x

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