What is the integral of sqrt(2+\sqrt{x)} ?

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

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integral of sqrt(2+\sqrt{x)}

\int 2 + x d x

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

Solution steps

\int 2 + x d x

Apply u-substitution

: \int 2 + u \cdot 2 u d u

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

Add a constant to the solution

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

x \to 1 lim (a x^{2} + b x + c)

x \to + 0 + lim (\frac{1}{x ^{\frac{2}{3}}})

x^{2} + 3

\int_{0}^{6} 2 π (6 - x) (6 x - x^{2}) d x

\int 3 e^{x} + 2 sin (x) d x

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