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

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

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

\int 1 + x d x

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

Solution steps

\int 1 + x d x

Apply u-substitution

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

= \int u \cdot 2 (u - 1) d u

Take the constant out:

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

= 2 \cdot \int u (u - 1) d u

Expand

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

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

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

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

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

Substitute back

u = 1 + x

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

Add a constant to the solution

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

\int x^{5} (1 + x^{3})^{\frac{3}{2}} d x

y^{^{'}} = 8 y^{2} + x y^{2}, y (0) = 1

4 (sin (t) y^{^{'}} + cos (t) y) = cos (t) sin^{5} (t)

\frac{2 x - 1}{x + 3}, at x = 4

\int \frac{2 x ^{3}}{36 x ^{2} + 1} d x

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