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

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

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integral of 3xsqrt(2x+3)

\int 3 x 2 x + 3 d x

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

Solution steps

\int 3 x 2 x + 3 d x

Take the constant out:

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

= 3 \cdot \int x 2 x + 3 d x

Apply u-substitution

: \int \frac{u ( u - 3 )}{4} d u

= 3 \cdot \int \frac{u ( u - 3 )}{4} d u

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int u (u - 3) d u

Expand

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

= 3 \cdot \frac{1}{4} \cdot \int u^{\frac{3}{2}} - 3 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

= 3 \cdot \frac{1}{4} (\int u^{\frac{3}{2}} d u - \int 3 u d u)

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

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

= 3 \cdot \frac{1}{4} (\frac{2}{5} u^{\frac{5}{2}} - 2 u^{\frac{3}{2}})

Substitute back

u = 2 x + 3

= 3 \cdot \frac{1}{4} (\frac{2}{5} (2 x + 3)^{\frac{5}{2}} - 2 (2 x + 3)^{\frac{3}{2}})

Simplify

3 \cdot \frac{1}{4} (\frac{2}{5} (2 x + 3)^{\frac{5}{2}} - 2 (2 x + 3)^{\frac{3}{2}}) : \frac{3}{5} (x - 1) (2 x + 3)^{\frac{3}{2}}

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

Add a constant to the solution

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

\int cos^{3} (\frac{x}{8}) d x

y = \frac{1}{x ^{5}}, (3, \frac{1}{243})

\int_{- 3}^{3} 18 - 2 x^{2} d x

x \to 2 - lim (\frac{3 - x}{2 - x})

(x y + y^{2} + x^{2}) d x - x^{2} d y = 0

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