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

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

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integral of x^2sqrt(2+x)

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

= \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

= \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}}

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

Substitute back

u = 2 + x

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

Add a constant to the solution

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

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