What is the integral of y^2sqrt(y+3) ?

The integral of y^2sqrt(y+3) is 2/7 (y+3)^{7/2}-12/5 (y+3)^{5/2}+6(y+3)^{3/2}+C

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integral of y^2sqrt(y+3)

\int y^{2} y + 3 d y

\frac{2}{7} (y + 3)^{\frac{7}{2}} - \frac{12}{5} (y + 3)^{\frac{5}{2}} + 6 (y + 3)^{\frac{3}{2}} + C

Solution steps

\int y^{2} y + 3 d y

Apply u-substitution

: \int (u - 3)^{2} u d u

= \int (u - 3)^{2} u d u

Expand

(u - 3)^{2} u : u^{\frac{5}{2}} - 6 u^{\frac{3}{2}} + 9 u

= \int u^{\frac{5}{2}} - 6 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{5}{2}} d u - \int 6 u^{\frac{3}{2}} d u + \int 9 u d u

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

\int 6 u^{\frac{3}{2}} d u = \frac{12}{5} u^{\frac{5}{2}}

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

= \frac{2}{7} u^{\frac{7}{2}} - \frac{12}{5} u^{\frac{5}{2}} + 6 u^{\frac{3}{2}}

Substitute back

u = y + 3

= \frac{2}{7} (y + 3)^{\frac{7}{2}} - \frac{12}{5} (y + 3)^{\frac{5}{2}} + 6 (y + 3)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (y + 3)^{\frac{7}{2}} - \frac{12}{5} (y + 3)^{\frac{5}{2}} + 6 (y + 3)^{\frac{3}{2}} + C

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What is the integral of y^2sqrt(y+3) ?
The integral of y^2sqrt(y+3) is 2/7 (y+3)^{7/2}-12/5 (y+3)^{5/2}+6(y+3)^{3/2}+C