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

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

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integral of xsqrt(1+3x)

\int x 1 + 3 x d x

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

Solution steps

\int x 1 + 3 x 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 = 1 + 3 x

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x - 2 y^{2} - 2 z^{2})

\int_{1}^{9} \frac{x - 1}{x} d x

x \to 2 lim (3 x^{2} - 4 x + 2)

x^{3} (x - 3)^{2}

\frac{d}{d x} (\frac{x}{x ^{3} - 1})

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