What is the integral of (sqrt(2y)+1)^2 ?

The integral of (sqrt(2y)+1)^2 is 1/3 (3y^2+4sqrt(2)y^{3/2}+3y)+C

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integral of (sqrt(2y)+1)^2

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

\frac{1}{3} (3 y^{2} + 42 y^{\frac{3}{2}} + 3 y) + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int (u + 1)^{2} d u

Expand

(u + 1)^{2} : u + 2 u + 1

= \frac{1}{2} \cdot \int u + 2 u + 1 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

= \frac{1}{2} (\int u d u + \int 2 u d u + \int 1 d u)

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

\int 2 u d u = \frac{4}{3} u^{\frac{3}{2}}

\int 1 d u = u

= \frac{1}{2} (\frac{u ^{2}}{2} + \frac{4}{3} u^{\frac{3}{2}} + u)

Substitute back

u = 2 y

= \frac{1}{2} (\frac{( 2 y ) ^{2}}{2} + \frac{4}{3} (2 y)^{\frac{3}{2}} + 2 y)

Simplify

\frac{1}{2} (\frac{( 2 y ) ^{2}}{2} + \frac{4}{3} (2 y)^{\frac{3}{2}} + 2 y) : \frac{1}{3} (3 y^{2} + 42 y^{\frac{3}{2}} + 3 y)

= \frac{1}{3} (3 y^{2} + 42 y^{\frac{3}{2}} + 3 y)

Add a constant to the solution

= \frac{1}{3} (3 y^{2} + 42 y^{\frac{3}{2}} + 3 y) + C

What is the integral of (sqrt(2y)+1)^2 ?
The integral of (sqrt(2y)+1)^2 is 1/3 (3y^2+4sqrt(2)y^{3/2}+3y)+C