What is the integral of sqrt(\sqrt{x)-1} ?

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

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integral of sqrt(\sqrt{x)-1}

\int x - 1 d x

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

Solution steps

\int x - 1 d x

Apply u-substitution

: \int u - 1 \cdot 2 u d u

= \int u - 1 \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 - 1 u d u

Apply u-substitution

: \int (v + 1) v d v

= 2 \cdot \int (v + 1) v d v

Expand

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

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

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

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

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

Substitute back

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

Add a constant to the solution

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

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

y^{^{''}} + 225 y = sec (15 x)

\frac{3 s e ^{- 2 s}}{5 s ^{2} + 45}

\int 4 t^{4} e^{2 t^{5}} d t

y^{^{''}} + 4 y = cos (2 x)

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