What is the integral of \sqrt[7]{x^3}(1+sqrt(x)) ?

The integral of \sqrt[7]{x^3}(1+sqrt(x)) is 1/3 (21/10 x^{10/7}+14/9 x^{27/14})+C

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integral of \sqrt[7]{x^3}(1+sqrt(x))

\frac{1}{3} (\frac{21}{10} x^{\frac{10}{7}} + \frac{14}{9} x^{\frac{27}{14}}) + C

Solution steps

Apply u-substitution

Take the constant out:

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

Expand

= \frac{1}{3} \cdot \int \frac{1}{u ^{\frac{11}{21}}} + \frac{1}{u ^{\frac{5}{14}}} 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

= \frac{1}{3} (\int \frac{1}{u ^{\frac{11}{21}}} d u + \int \frac{1}{u ^{\frac{5}{14}}} d u)

\int \frac{1}{u ^{\frac{11}{21}}} d u = \frac{21}{10} u^{\frac{10}{21}}

\int \frac{1}{u ^{\frac{5}{14}}} d u = \frac{14}{9} u^{\frac{9}{14}}

= \frac{1}{3} (\frac{21}{10} u^{\frac{10}{21}} + \frac{14}{9} u^{\frac{9}{14}})

Substitute back

u = x^{3}

= \frac{1}{3} (\frac{21}{10} (x^{3})^{\frac{10}{21}} + \frac{14}{9} (x^{3})^{\frac{9}{14}})

Simplify

\frac{1}{3} (\frac{21}{10} (x^{3})^{\frac{10}{21}} + \frac{14}{9} (x^{3})^{\frac{9}{14}}) : \frac{1}{3} (\frac{21}{10} x^{\frac{10}{7}} + \frac{14}{9} x^{\frac{27}{14}})

= \frac{1}{3} (\frac{21}{10} x^{\frac{10}{7}} + \frac{14}{9} x^{\frac{27}{14}})

Add a constant to the solution

= \frac{1}{3} (\frac{21}{10} x^{\frac{10}{7}} + \frac{14}{9} x^{\frac{27}{14}}) + C

x \to 1.9 lim (- x^{2} + 5 x - 2)

x \to - 5 lim (\frac{- 6 x}{x + 5})

\frac{d y}{d t} = - 3 (y - 2)

\frac{\partial}{\partial x} (3 x + 2 x y^{2})

\frac{d}{d x} (2 ln (1 + x^{2}) - 3^{x})

What is the integral of \sqrt[7]{x^3}(1+sqrt(x)) ?
The integral of \sqrt[7]{x^3}(1+sqrt(x)) is 1/3 (21/10 x^{10/7}+14/9 x^{27/14})+C