What is the integral of 8x^2\sqrt[4]{4+4x^3} ?

The integral of 8x^2\sqrt[4]{4+4x^3} is 8/15 (4x^3+4)^{5/4}+C

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integral of 8x^2\sqrt[4]{4+4x^3}

\frac{8}{15} (4 x^{3} + 4)^{\frac{5}{4}} + C

Solution steps

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

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

= 8 \cdot \frac{1}{12} \cdot \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = 4 + 4 x^{3}

= 8 \cdot \frac{1}{12} \cdot \frac{4}{5} (4 + 4 x^{3})^{\frac{5}{4}}

Simplify

8 \cdot \frac{1}{12} \cdot \frac{4}{5} (4 + 4 x^{3})^{\frac{5}{4}} : \frac{8}{15} (4 x^{3} + 4)^{\frac{5}{4}}

= \frac{8}{15} (4 x^{3} + 4)^{\frac{5}{4}}

Add a constant to the solution

= \frac{8}{15} (4 x^{3} + 4)^{\frac{5}{4}} + C

\int \frac{( e ^{2 y} - y )}{e ^{y}} d y

\frac{1}{s ^{2} ( s ^{2} + 4 )}

(x^{2} - 12) (2 x - 9)

\int_{x}^{2} 0.5 d y

\int (sin (x))^{5} d x

What is the integral of 8x^2\sqrt[4]{4+4x^3} ?
The integral of 8x^2\sqrt[4]{4+4x^3} is 8/15 (4x^3+4)^{5/4}+C