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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = 3 + x^{4}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (∣ x^{2} - 6 x + 5 ∣)

\int_{0}^{1} \frac{1}{x ^{5}} d x

5 t \frac{d y}{d t} + y = t^{4}

h \to 0 lim (\frac{ln ( e + h ) - 1}{h})

\frac{d}{d x} (3 x^{2} + \frac{2}{x} - \frac{1}{x ^{2}})

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