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

The integral of x^2sqrt(x+3) is 2/7 (x+3)^{7/2}-12/5 (x+3)^{5/2}+6(x+3)^{3/2}+C

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integral of x^2sqrt(x+3)

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

= \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

= \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{2}{7} u^{\frac{7}{2}} - \frac{12}{5} u^{\frac{5}{2}} + 6 u^{\frac{3}{2}}

Substitute back

u = x + 3

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

Add a constant to the solution

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

cos (2 x)

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

\frac{d}{d x} (\frac{7}{x ^{4}})

x \to 0 lim (- 2 csc (2 x))

\frac{\partial}{\partial x} (- sin (x))

What is the integral of x^2sqrt(x+3) ?
The integral of x^2sqrt(x+3) is 2/7 (x+3)^{7/2}-12/5 (x+3)^{5/2}+6(x+3)^{3/2}+C