What is the integral of sqrt(10+\sqrt{x)} ?

The integral of sqrt(10+\sqrt{x)} is 2(2/5 (10+sqrt(x))^{5/2}-20/3 (10+sqrt(x))^{3/2})+C

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integral of sqrt(10+\sqrt{x)}

\int 10 + x d x

2 (\frac{2}{5} (10 + x)^{\frac{5}{2}} - \frac{20}{3} (10 + x)^{\frac{3}{2}}) + C

Solution steps

\int 10 + x d x

Apply u-substitution

: \int 10 + u \cdot 2 u d u

= \int 10 + 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 10 + u u d u

Apply u-substitution

: \int (v - 10) v d v

= 2 \cdot \int (v - 10) v d v

Expand

(v - 10) v : v^{\frac{3}{2}} - 10 v

= 2 \cdot \int v^{\frac{3}{2}} - 10 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 10 v d v)

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

\int 10 v d v = \frac{20}{3} v^{\frac{3}{2}}

= 2 (\frac{2}{5} v^{\frac{5}{2}} - \frac{20}{3} v^{\frac{3}{2}})

Substitute back

= 2 (\frac{2}{5} (10 + x)^{\frac{5}{2}} - \frac{20}{3} (10 + x)^{\frac{3}{2}})

Add a constant to the solution

= 2 (\frac{2}{5} (10 + x)^{\frac{5}{2}} - \frac{20}{3} (10 + x)^{\frac{3}{2}}) + C

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What is the integral of sqrt(10+\sqrt{x)} ?
The integral of sqrt(10+\sqrt{x)} is 2(2/5 (10+sqrt(x))^{5/2}-20/3 (10+sqrt(x))^{3/2})+C