What is the integral of 4x^4(27+x^5)^{1/3} ?

The integral of 4x^4(27+x^5)^{1/3} is 3/5 (27+x^5)^{4/3}+C

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integral of 4x^4(27+x^5)^{1/3}

\int 4 x^{4} (27 + x^{5})^{\frac{1}{3}} d x

\frac{3}{5} (27 + x^{5})^{\frac{4}{3}} + C

Solution steps

\int 4 x^{4} (27 + x^{5})^{\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^{4} (27 + x^{5})^{\frac{1}{3}} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{3}}}{5} d u

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 27 + x^{5}

= 4 \cdot \frac{1}{5} \cdot \frac{3}{4} (27 + x^{5})^{\frac{4}{3}}

Simplify

4 \cdot \frac{1}{5} \cdot \frac{3}{4} (27 + x^{5})^{\frac{4}{3}} : \frac{3}{5} (27 + x^{5})^{\frac{4}{3}}

= \frac{3}{5} (27 + x^{5})^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{5} (27 + x^{5})^{\frac{4}{3}} + C

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x \to - \infty lim (4 x^{2})

What is the integral of 4x^4(27+x^5)^{1/3} ?
The integral of 4x^4(27+x^5)^{1/3} is 3/5 (27+x^5)^{4/3}+C