What is the integral of sqrt(\sqrt{y+1)-2} ?

The integral of sqrt(\sqrt{y+1)-2} is 2(2/5 (sqrt(y+1)-2)^{5/2}+4/3 (sqrt(y+1)-2)^{3/2})+C

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integral of sqrt(\sqrt{y+1)-2}

\int y + 1 - 2 d y

2 (\frac{2}{5} (y + 1 - 2)^{\frac{5}{2}} + \frac{4}{3} (y + 1 - 2)^{\frac{3}{2}}) + C

Solution steps

\int y + 1 - 2 d y

Apply u-substitution

: \int u - 2 \cdot 2 u d u

= \int u - 2 \cdot 2 u d u

Take the constant out:

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

= 2 \cdot \int u - 2 u d u

Apply u-substitution

: \int (v + 2) v d v

= 2 \cdot \int (v + 2) v d v

Expand

(v + 2) v : v^{\frac{3}{2}} + 2 v

= 2 \cdot \int v^{\frac{3}{2}} + 2 v d v

Apply the Sum Rule:

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

= 2 (\int v^{\frac{3}{2}} d v + \int 2 v d v)

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

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

= 2 (\frac{2}{5} v^{\frac{5}{2}} + \frac{4}{3} v^{\frac{3}{2}})

Substitute back

= 2 (\frac{2}{5} (y + 1 - 2)^{\frac{5}{2}} + \frac{4}{3} (y + 1 - 2)^{\frac{3}{2}})

Add a constant to the solution

= 2 (\frac{2}{5} (y + 1 - 2)^{\frac{5}{2}} + \frac{4}{3} (y + 1 - 2)^{\frac{3}{2}}) + C

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What is the integral of sqrt(\sqrt{y+1)-2} ?
The integral of sqrt(\sqrt{y+1)-2} is 2(2/5 (sqrt(y+1)-2)^{5/2}+4/3 (sqrt(y+1)-2)^{3/2})+C