What is the integral of (16x^4)/(\sqrt[3]{25x^5+4)} ?

The integral of (16x^4)/(\sqrt[3]{25x^5+4)} is 24/125 (25x^5+4)^{2/3}+C

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integral of (16x^4)/(\sqrt[3]{25x^5+4)}

\int \frac{16 x ^{4}}{3 25 x ^{5} + 4} d x

\frac{24}{125} (25 x^{5} + 4)^{\frac{2}{3}} + C

Solution steps

\int \frac{16 x ^{4}}{3 25 x ^{5} + 4} d x

Take the constant out:

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

= 16 \cdot \int \frac{x ^{4}}{3 25 x ^{5} + 4} d x

Apply u-substitution

: \int \frac{1}{125 3 u} d u

= 16 \cdot \int \frac{1}{125 3 u} d u

Take the constant out:

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

= 16 \cdot \frac{1}{125} \cdot \int \frac{1}{3 u} d u

Apply the Power Rule

: \frac{3}{2} u^{\frac{2}{3}}

= 16 \cdot \frac{1}{125} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 25 x^{5} + 4

= 16 \cdot \frac{1}{125} \cdot \frac{3}{2} (25 x^{5} + 4)^{\frac{2}{3}}

Simplify

16 \cdot \frac{1}{125} \cdot \frac{3}{2} (25 x^{5} + 4)^{\frac{2}{3}} : \frac{24}{125} (25 x^{5} + 4)^{\frac{2}{3}}

= \frac{24}{125} (25 x^{5} + 4)^{\frac{2}{3}}

Add a constant to the solution

= \frac{24}{125} (25 x^{5} + 4)^{\frac{2}{3}} + C

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What is the integral of (16x^4)/(\sqrt[3]{25x^5+4)} ?
The integral of (16x^4)/(\sqrt[3]{25x^5+4)} is 24/125 (25x^5+4)^{2/3}+C