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 4 + u \cdot 2 u d u

= \int 4 + u \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 4 + u u d u

Apply u-substitution

: \int (v - 4) v d v

= 2 \cdot \int (v - 4) v d v

Expand

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

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

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

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

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

Substitute back

= 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

y^{^{''}} + 3 y^{^{'}} + 2 y = 6

x \to 6 lim (\frac{1}{x - 2})

f (x) = cos (x), 0, 2 π

y = - 2 - 7 x^{2}, (- 2, - 30)

\int \frac{3 cos ( x )}{sin ^{2} ( x ) + sin ( x )} 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