What is the integral of xsqrt(9+x) ?

The integral of xsqrt(9+x) is 2/5 (9+x)^{5/2}-6(9+x)^{3/2}+C

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integral of xsqrt(9+x)

\int x 9 + x d x

\frac{2}{5} (9 + x)^{\frac{5}{2}} - 6 (9 + x)^{\frac{3}{2}} + C

Solution steps

\int x 9 + x d x

Apply u-substitution

: \int (u - 9) u d u

= \int (u - 9) u d u

Expand

(u - 9) u : u^{\frac{3}{2}} - 9 u

= \int u^{\frac{3}{2}} - 9 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{3}{2}} d u - \int 9 u d u

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

\int 9 u d u = 6 u^{\frac{3}{2}}

= \frac{2}{5} u^{\frac{5}{2}} - 6 u^{\frac{3}{2}}

Substitute back

u = 9 + x

= \frac{2}{5} (9 + x)^{\frac{5}{2}} - 6 (9 + x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} (9 + x)^{\frac{5}{2}} - 6 (9 + x)^{\frac{3}{2}} + C

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What is the integral of xsqrt(9+x) ?
The integral of xsqrt(9+x) is 2/5 (9+x)^{5/2}-6(9+x)^{3/2}+C