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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

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

Substitute back

u = 4 + x^{4}

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

Simplify

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

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

Add a constant to the solution

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

\frac{r}{r ^{2} + 4}

x \frac{d y}{d x} = x^{3} + 4 x^{3} y

n = 1 \sum \infty \frac{n 5 ^{n}}{n !}

5 tan (x)

x (x - 14)

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