What is the integral of 6x^2sqrt(x+2) ?

The integral of 6x^2sqrt(x+2) is 6(2/7 (x+2)^{7/2}-8/5 (x+2)^{5/2}+8/3 (x+2)^{3/2})+C

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integral of 6x^2sqrt(x+2)

\int 6 x^{2} x + 2 d x

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

Solution steps

\int 6 x^{2} x + 2 d x

Take the constant out:

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

= 6 \cdot \int x^{2} x + 2 d x

Apply u-substitution

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

= 6 \cdot \int (u - 2)^{2} u d u

Expand

(u - 2)^{2} u : u^{\frac{5}{2}} - 4 u^{\frac{3}{2}} + 4 u

= 6 \cdot \int u^{\frac{5}{2}} - 4 u^{\frac{3}{2}} + 4 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 (\int u^{\frac{5}{2}} d u - \int 4 u^{\frac{3}{2}} d u + \int 4 u d u)

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

\int 4 u^{\frac{3}{2}} d u = \frac{8}{5} u^{\frac{5}{2}}

\int 4 u d u = \frac{8}{3} u^{\frac{3}{2}}

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

Substitute back

u = x + 2

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

Add a constant to the solution

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

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What is the integral of 6x^2sqrt(x+2) ?
The integral of 6x^2sqrt(x+2) is 6(2/7 (x+2)^{7/2}-8/5 (x+2)^{5/2}+8/3 (x+2)^{3/2})+C