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

The integral of x^2sqrt(2x+3) is 1/35 (5x^2-6x+6)(2x+3)^{3/2}+C

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

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

\frac{1}{35} (5 x^{2} - 6 x + 6) (2 x + 3)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ( u - 3 ) ^{2}}{8} d u

= \int \frac{u ( u - 3 ) ^{2}}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u (u - 3)^{2} d u

Expand

u (u - 3)^{2} : u^{\frac{5}{2}} - 6 u^{\frac{3}{2}} + 9 u

= \frac{1}{8} \cdot \int u^{\frac{5}{2}} - 6 u^{\frac{3}{2}} + 9 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}{8} (\int u^{\frac{5}{2}} d u - \int 6 u^{\frac{3}{2}} d u + \int 9 u d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 6 u^{\frac{3}{2}} d u = \frac{12}{5} u^{\frac{5}{2}}

\int 9 u d u = 6 u^{\frac{3}{2}}

= \frac{1}{8} (\frac{2}{7} u^{\frac{7}{2}} - \frac{12}{5} u^{\frac{5}{2}} + 6 u^{\frac{3}{2}})

Substitute back

u = 2 x + 3

= \frac{1}{8} (\frac{2}{7} (2 x + 3)^{\frac{7}{2}} - \frac{12}{5} (2 x + 3)^{\frac{5}{2}} + 6 (2 x + 3)^{\frac{3}{2}})

Simplify

\frac{1}{8} (\frac{2}{7} (2 x + 3)^{\frac{7}{2}} - \frac{12}{5} (2 x + 3)^{\frac{5}{2}} + 6 (2 x + 3)^{\frac{3}{2}}) : \frac{1}{35} (5 x^{2} - 6 x + 6) (2 x + 3)^{\frac{3}{2}}

= \frac{1}{35} (5 x^{2} - 6 x + 6) (2 x + 3)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{35} (5 x^{2} - 6 x + 6) (2 x + 3)^{\frac{3}{2}} + C

f (x) = 3 arccos (\frac{x}{2})

\frac{d}{d x} (\frac{3 x}{5 x ^{2} - 2 x - 9})

y = x^{2} - 18, y = 17 x

\int_{e}^{e^{8}} \frac{1}{x ln ( x ^{7} )} d x

\frac{d y}{d x} = e^{- x} + y

What is the integral of x^2sqrt(2x+3) ?
The integral of x^2sqrt(2x+3) is 1/35 (5x^2-6x+6)(2x+3)^{3/2}+C