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

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

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integral of (x^3)/(\sqrt[3]{8+x^4)}

\int \frac{x ^{3}}{3 8 + x ^{4}} d x

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

Solution steps

\int \frac{x ^{3}}{3 8 + x ^{4}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

= \frac{1}{4} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 8 + x^{4}

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

Simplify

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

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

Add a constant to the solution

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

\frac{9}{x}

\int (\frac{1}{x} - 2) d x

\int \frac{2 x}{x ^{4}} d x

\int \frac{1}{z ln ( z ) ln ( ln ( z ) )} d z

\int \frac{x}{( 1 + x ) ^{\frac{1}{2}}} d x

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