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

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

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integral of xsqrt(x+9)

\int x x + 9 d x

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

Solution steps

\int x x + 9 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 = x + 9

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

Add a constant to the solution

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

x \to - 1 lim (x^{2} - 5 x + 6)

\frac{\partial}{\partial y} (3 x^{2} - 3 y)

\int \frac{( x ^{2} - 9 )}{x ^{3}} d x

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

\int \frac{tan ( x ) + 1}{tan ( x ) - 1} d x

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