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

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

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integral of \sqrt[3]{x+4}

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

Solution steps

Apply u-substitution

Apply the Power Rule

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

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

Substitute back

u = x + 4

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

Add a constant to the solution

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

x \to 5 lim (\frac{x ^{2} - 25}{3 x - 15})

(1 - x)^{2} + x^{3}

x \to 5 lim (\frac{1}{x ^{2} + x})

\int \frac{1}{- 2 x ^{2} + 8 x + 4} d x

\int_{0}^{3} (2 x - x^{2})^{3} (1 - x) d x

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