What is the integral of x^2sqrt(9+x) ?

The integral of x^2sqrt(9+x) is 2/7 (9+x)^{7/2}-36/5 (9+x)^{5/2}+54(9+x)^{3/2}+C

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integral of x^2sqrt(9+x)

\int x^{2} 9 + x d x

\frac{2}{7} (9 + x)^{\frac{7}{2}} - \frac{36}{5} (9 + x)^{\frac{5}{2}} + 54 (9 + x)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 9 + x d x

Apply u-substitution

: \int (u - 9)^{2} u d u

= \int (u - 9)^{2} u d u

Expand

(u - 9)^{2} u : u^{\frac{5}{2}} - 18 u^{\frac{3}{2}} + 81 u

= \int u^{\frac{5}{2}} - 18 u^{\frac{3}{2}} + 81 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 18 u^{\frac{3}{2}} d u + \int 81 u d u

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

\int 18 u^{\frac{3}{2}} d u = \frac{36}{5} u^{\frac{5}{2}}

\int 81 u d u = 54 u^{\frac{3}{2}}

= \frac{2}{7} u^{\frac{7}{2}} - \frac{36}{5} u^{\frac{5}{2}} + 54 u^{\frac{3}{2}}

Substitute back

u = 9 + x

= \frac{2}{7} (9 + x)^{\frac{7}{2}} - \frac{36}{5} (9 + x)^{\frac{5}{2}} + 54 (9 + x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (9 + x)^{\frac{7}{2}} - \frac{36}{5} (9 + x)^{\frac{5}{2}} + 54 (9 + x)^{\frac{3}{2}} + C

\frac{x}{x ^{2} + 1}

(x^{2} + 1) \frac{d y}{d x} + 6 x (y - 1) = 0, y (0) = 8

\int (x + 6)^{4} d x

y = x, y = x, x = 0, x = 2

f (x) = x x^{2} + 1

What is the integral of x^2sqrt(9+x) ?
The integral of x^2sqrt(9+x) is 2/7 (9+x)^{7/2}-36/5 (9+x)^{5/2}+54(9+x)^{3/2}+C