What is the integral of x^2sqrt(4+x) ?

The integral of x^2sqrt(4+x) is 2/7 (4+x)^{7/2}-16/5 (4+x)^{5/2}+32/3 (4+x)^{3/2}+C

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integral of x^2sqrt(4+x)

\int x^{2} 4 + x d x

\frac{2}{7} (4 + x)^{\frac{7}{2}} - \frac{16}{5} (4 + x)^{\frac{5}{2}} + \frac{32}{3} (4 + x)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 4 + x d x

Apply u-substitution

: \int (u - 4)^{2} u d u

= \int (u - 4)^{2} u d u

Expand

(u - 4)^{2} u : u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 u

= \int u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 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 8 u^{\frac{3}{2}} d u + \int 16 u d u

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

\int 8 u^{\frac{3}{2}} d u = \frac{16}{5} u^{\frac{5}{2}}

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

= \frac{2}{7} u^{\frac{7}{2}} - \frac{16}{5} u^{\frac{5}{2}} + \frac{32}{3} u^{\frac{3}{2}}

Substitute back

u = 4 + x

= \frac{2}{7} (4 + x)^{\frac{7}{2}} - \frac{16}{5} (4 + x)^{\frac{5}{2}} + \frac{32}{3} (4 + x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (4 + x)^{\frac{7}{2}} - \frac{16}{5} (4 + x)^{\frac{5}{2}} + \frac{32}{3} (4 + x)^{\frac{3}{2}} + C

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What is the integral of x^2sqrt(4+x) ?
The integral of x^2sqrt(4+x) is 2/7 (4+x)^{7/2}-16/5 (4+x)^{5/2}+32/3 (4+x)^{3/2}+C