What is the integral of x^2sqrt(8+x) ?

The integral of x^2sqrt(8+x) is 2/7 (8+x)^{7/2}-32/5 (8+x)^{5/2}+128/3 (8+x)^{3/2}+C

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integral of x^2sqrt(8+x)

\int x^{2} 8 + x d x

\frac{2}{7} (8 + x)^{\frac{7}{2}} - \frac{32}{5} (8 + x)^{\frac{5}{2}} + \frac{128}{3} (8 + x)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 8 + x d x

Apply u-substitution

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

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

Expand

(u - 8)^{2} u : u^{\frac{5}{2}} - 16 u^{\frac{3}{2}} + 64 u

= \int u^{\frac{5}{2}} - 16 u^{\frac{3}{2}} + 64 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 16 u^{\frac{3}{2}} d u + \int 64 u d u

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

\int 16 u^{\frac{3}{2}} d u = \frac{32}{5} u^{\frac{5}{2}}

\int 64 u d u = \frac{128}{3} u^{\frac{3}{2}}

= \frac{2}{7} u^{\frac{7}{2}} - \frac{32}{5} u^{\frac{5}{2}} + \frac{128}{3} u^{\frac{3}{2}}

Substitute back

u = 8 + x

= \frac{2}{7} (8 + x)^{\frac{7}{2}} - \frac{32}{5} (8 + x)^{\frac{5}{2}} + \frac{128}{3} (8 + x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (8 + x)^{\frac{7}{2}} - \frac{32}{5} (8 + x)^{\frac{5}{2}} + \frac{128}{3} (8 + x)^{\frac{3}{2}} + C

\frac{d}{d x} (arccot (\frac{1}{2}) (x - \frac{1}{x}))

\int \frac{7}{9 + 4 x ^{2}} d x

\frac{d}{d x} (ln (\frac{x}{7}))

y^{^{''}} + 5 y^{^{'}} + 6 = 1

3 e^{- 4 t}

What is the integral of x^2sqrt(8+x) ?
The integral of x^2sqrt(8+x) is 2/7 (8+x)^{7/2}-32/5 (8+x)^{5/2}+128/3 (8+x)^{3/2}+C