What is the integral of sqrt(4-\sqrt{x)} ?

The integral of sqrt(4-\sqrt{x)} is -2(-2/5 (4-sqrt(x))^{5/2}+8/3 (4-sqrt(x))^{3/2})+C

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integral of sqrt(4-\sqrt{x)}

\int 4 - x d x

- 2 (- \frac{2}{5} (4 - x)^{\frac{5}{2}} + \frac{8}{3} (4 - x)^{\frac{3}{2}}) + C

Solution steps

\int 4 - x d x

Apply u-substitution

: \int - 2 (- u + 4) u d u

= \int - 2 (- u + 4) 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 + 4) u d u

Expand

(- u + 4) u : - u^{\frac{3}{2}} + 4 u

= - 2 \cdot \int - u^{\frac{3}{2}} + 4 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

= - 2 (- \int u^{\frac{3}{2}} d u + \int 4 u d u)

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

\int 4 u d u = \frac{8}{3} u^{\frac{3}{2}}

= - 2 (- \frac{2}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}})

Substitute back

u = 4 - x

= - 2 (- \frac{2}{5} (4 - x)^{\frac{5}{2}} + \frac{8}{3} (4 - x)^{\frac{3}{2}})

Add a constant to the solution

= - 2 (- \frac{2}{5} (4 - x)^{\frac{5}{2}} + \frac{8}{3} (4 - x)^{\frac{3}{2}}) + C

x \to - 10 lim (\frac{1}{x + 10})

x \to 0 - lim (e^{x})

\int_{0}^{2 π} cos (t) sin (t) d t

\int \frac{3}{2} (x + 1)^{\frac{1}{2}} d x

What is the integral of sqrt(4-\sqrt{x)} ?
The integral of sqrt(4-\sqrt{x)} is -2(-2/5 (4-sqrt(x))^{5/2}+8/3 (4-sqrt(x))^{3/2})+C