What is the integral of 5x^3\sqrt[3]{(2x^4-4)^2} ?

The integral of 5x^3\sqrt[3]{(2x^4-4)^2} is 3/8 (2x^4-4)^{5/3}+C

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integral of 5x^3\sqrt[3]{(2x^4-4)^2}

\frac{3}{8} (2 x^{4} - 4)^{\frac{5}{3}} + C

Solution steps

Take the constant out:

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

Apply u-substitution

: \int \frac{u ^{\frac{2}{3}}}{8} d u

= 5 \cdot \int \frac{u ^{\frac{2}{3}}}{8} d u

Take the constant out:

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

= 5 \cdot \frac{1}{8} \cdot \int u^{\frac{2}{3}} d u

Apply the Power Rule

: \frac{3}{5} u^{\frac{5}{3}}

= 5 \cdot \frac{1}{8} \cdot \frac{3}{5} u^{\frac{5}{3}}

Substitute back

u = 2 x^{4} - 4

= 5 \cdot \frac{1}{8} \cdot \frac{3}{5} (2 x^{4} - 4)^{\frac{5}{3}}

Simplify

5 \cdot \frac{1}{8} \cdot \frac{3}{5} (2 x^{4} - 4)^{\frac{5}{3}} : \frac{3}{8} (2 x^{4} - 4)^{\frac{5}{3}}

= \frac{3}{8} (2 x^{4} - 4)^{\frac{5}{3}}

Add a constant to the solution

= \frac{3}{8} (2 x^{4} - 4)^{\frac{5}{3}} + C

\frac{\partial}{\partial x} (tan (x y))

y = e^{t a n (x)}

x \to 0 lim (\frac{cos ( x )}{96})

\frac{d y}{d t} + 0.3 t y = 9 t

6 t - 3

What is the integral of 5x^3\sqrt[3]{(2x^4-4)^2} ?
The integral of 5x^3\sqrt[3]{(2x^4-4)^2} is 3/8 (2x^4-4)^{5/3}+C