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

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

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integral of sqrt(1+\sqrt{2x)}

\int 1 + 2 x d x

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

Solution steps

\int 1 + 2 x d x

Apply u-substitution

: \int \frac{1 + u}{2} d u

= \int \frac{1 + u}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 + u d u

Apply u-substitution

: \int v \cdot 2 (v - 1) d v

= \frac{1}{2} \cdot \int v \cdot 2 (v - 1) d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int v (v - 1) d v

Expand

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

= \frac{1}{2} \cdot 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

= \frac{1}{2} \cdot 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}}

= \frac{1}{2} \cdot 2 (\frac{2}{5} v^{\frac{5}{2}} - \frac{2}{3} v^{\frac{3}{2}})

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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