What is the integral of \sqrt[3]{1+sqrt(x)} ?

The integral of \sqrt[3]{1+sqrt(x)} is 2(3/7 (1+sqrt(x))^{7/3}-3/4 (1+sqrt(x))^{4/3})+C

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integral of \sqrt[3]{1+sqrt(x)}

\int 3 1 + x d x

2 (\frac{3}{7} (1 + x)^{\frac{7}{3}} - \frac{3}{4} (1 + x)^{\frac{4}{3}}) + C

Solution steps

\int 3 1 + x d x

Apply u-substitution

: \int 3 1 + u \cdot 2 u d u

= \int 3 1 + 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 3 1 + u u d u

Apply u-substitution

: \int (v - 1) 3 v d v

= 2 \cdot \int (v - 1) 3 v d v

Expand

(v - 1) 3 v : v^{\frac{4}{3}} - 3 v

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

\int v^{\frac{4}{3}} d v = \frac{3}{7} v^{\frac{7}{3}}

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

= 2 (\frac{3}{7} v^{\frac{7}{3}} - \frac{3}{4} v^{\frac{4}{3}})

Substitute back

= 2 (\frac{3}{7} (1 + x)^{\frac{7}{3}} - \frac{3}{4} (1 + x)^{\frac{4}{3}})

Add a constant to the solution

= 2 (\frac{3}{7} (1 + x)^{\frac{7}{3}} - \frac{3}{4} (1 + x)^{\frac{4}{3}}) + C

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What is the integral of \sqrt[3]{1+sqrt(x)} ?
The integral of \sqrt[3]{1+sqrt(x)} is 2(3/7 (1+sqrt(x))^{7/3}-3/4 (1+sqrt(x))^{4/3})+C