What is the integral of 6xsqrt(3x+1) ?

The integral of 6xsqrt(3x+1) is 2/3 (2/5 (3x+1)^{5/2}-2/3 (3x+1)^{3/2})+C

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integral of 6xsqrt(3x+1)

\int 6 x 3 x + 1 d x

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

Solution steps

\int 6 x 3 x + 1 d x

Take the constant out:

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

= 6 \cdot \int x 3 x + 1 d x

Apply u-substitution

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

= 6 \cdot \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

= 6 \cdot \frac{1}{9} \cdot \int u (u - 1) d u

Expand

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

= 6 \cdot \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

= 6 \cdot \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}}

= 6 \cdot \frac{1}{9} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 3 x + 1

= 6 \cdot \frac{1}{9} (\frac{2}{5} (3 x + 1)^{\frac{5}{2}} - \frac{2}{3} (3 x + 1)^{\frac{3}{2}})

Simplify

6 \cdot \frac{1}{9} (\frac{2}{5} (3 x + 1)^{\frac{5}{2}} - \frac{2}{3} (3 x + 1)^{\frac{3}{2}}) : \frac{2}{3} (\frac{2}{5} (3 x + 1)^{\frac{5}{2}} - \frac{2}{3} (3 x + 1)^{\frac{3}{2}})

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

Add a constant to the solution

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

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

\frac{\partial}{\partial u ( v )} (\frac{e ^{v}}{u ( v ) + v ^{3}})

\int_{1}^{\infty} \frac{5}{x ^{2} + 1} d x

n = 2 \sum \infty \frac{( - 1 ) ^{n}}{n}

\int x^{3} e^{- 2 x} d x

What is the integral of 6xsqrt(3x+1) ?
The integral of 6xsqrt(3x+1) is 2/3 (2/5 (3x+1)^{5/2}-2/3 (3x+1)^{3/2})+C