What is the integral of x^2(1+x^3)^{1/4} ?

The integral of x^2(1+x^3)^{1/4} is 4/15 (1+x^3)^{5/4}+C

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integral of x^2(1+x^3)^{1/4}

\int x^{2} (1 + x^{3})^{\frac{1}{4}} d x

\frac{4}{15} (1 + x^{3})^{\frac{5}{4}} + C

Solution steps

\int x^{2} (1 + x^{3})^{\frac{1}{4}} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{4}}}{3} d u

= \int \frac{u ^{\frac{1}{4}}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u^{\frac{1}{4}} d u

Apply the Power Rule

: \frac{4}{5} u^{\frac{5}{4}}

= \frac{1}{3} \cdot \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = 1 + x^{3}

= \frac{1}{3} \cdot \frac{4}{5} (1 + x^{3})^{\frac{5}{4}}

Simplify

\frac{1}{3} \cdot \frac{4}{5} (1 + x^{3})^{\frac{5}{4}} : \frac{4}{15} (1 + x^{3})^{\frac{5}{4}}

= \frac{4}{15} (1 + x^{3})^{\frac{5}{4}}

Add a constant to the solution

= \frac{4}{15} (1 + x^{3})^{\frac{5}{4}} + C

\frac{\partial}{\partial x} (5 x y^{2})

f (x) = 2 x^{2} - 6 x^{\frac{1}{3}}

y = 3 x (- x + 1), y = - 3 x

\frac{d}{d x} (e^{m x})

7 + 2 x + x^{3} (5 x^{2} + 1)^{5}

What is the integral of x^2(1+x^3)^{1/4} ?
The integral of x^2(1+x^3)^{1/4} is 4/15 (1+x^3)^{5/4}+C