What is the integral of (1+y)y^{1/2} ?

The integral of (1+y)y^{1/2} is 2/3 y^{3/2}+2/5 y^{5/2}+C

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integral of (1+y)y^{1/2}

\int (1 + y) y^{\frac{1}{2}} d y

\frac{2}{3} y^{\frac{3}{2}} + \frac{2}{5} y^{\frac{5}{2}} + C

Solution steps

\int (1 + y) y^{\frac{1}{2}} d y

Expand

(1 + y) y^{\frac{1}{2}} : y^{\frac{1}{2}} + y^{\frac{3}{2}}

= \int y^{\frac{1}{2}} + y^{\frac{3}{2}} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int y^{\frac{1}{2}} d y + \int y^{\frac{3}{2}} d y

\int y^{\frac{1}{2}} d y = \frac{2}{3} y^{\frac{3}{2}}

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

= \frac{2}{3} y^{\frac{3}{2}} + \frac{2}{5} y^{\frac{5}{2}}

Add a constant to the solution

= \frac{2}{3} y^{\frac{3}{2}} + \frac{2}{5} y^{\frac{5}{2}} + C

What is the integral of (1+y)y^{1/2} ?
The integral of (1+y)y^{1/2} is 2/3 y^{3/2}+2/5 y^{5/2}+C