What is the integral of ysqrt(y+9) ?

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

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integral of ysqrt(y+9)

\int y y + 9 d y

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

Solution steps

\int y y + 9 d y

Apply u-substitution

= \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 = y + 9

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

Add a constant to the solution

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

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