What is the integral of sqrt(1+(4/(\sqrt[3]{x)))^2} ?

The integral of sqrt(1+(4/(\sqrt[3]{x)))^2} is (x^{2/3}+16)^{3/2}+C

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integral of sqrt(1+(4/(\sqrt[3]{x)))^2}

(x^{\frac{2}{3}} + 16)^{\frac{3}{2}} + C

Solution steps

Simplify

= \int \frac{x ^{\frac{2}{3}} + 16}{x ^{\frac{2}{3}}} d x

\frac{x ^{\frac{2}{3}} + 16}{x ^{\frac{2}{3}}} = \frac{16}{x ^{\frac{2}{3}}} + 1

= \int \frac{16}{x ^{\frac{2}{3}}} + 1 d x

Apply u-substitution

: \int \frac{3 u + 16}{2} d u

= \int \frac{3 u + 16}{2} d u

Take the constant out:

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

= \frac{3}{2} \cdot \int u + 16 d u

Apply u-substitution

: \int v d v

= \frac{3}{2} \cdot \int v d v

Apply the Power Rule

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

= \frac{3}{2} \cdot \frac{2}{3} v^{\frac{3}{2}}

Substitute back

= \frac{3}{2} \cdot \frac{2}{3} (x^{\frac{2}{3}} + 16)^{\frac{3}{2}}

Simplify

\frac{3}{2} \cdot \frac{2}{3} (x^{\frac{2}{3}} + 16)^{\frac{3}{2}} : (x^{\frac{2}{3}} + 16)^{\frac{3}{2}}

= (x^{\frac{2}{3}} + 16)^{\frac{3}{2}}

Add a constant to the solution

= (x^{\frac{2}{3}} + 16)^{\frac{3}{2}} + C

\frac{d}{d x} (3 (4 - 9 x)^{4})

\int \frac{7 e ^{7 x}}{1 + e ^{7 x}} d x

x \to 0 lim (\frac{tan ( 12 x )}{sin ( 4 x )})

\frac{\partial}{\partial x} (y e^{x} - x^{2} y^{- 3})

y = \frac{4 x}{x ^{2} + 1}, (0, 0)

What is the integral of sqrt(1+(4/(\sqrt[3]{x)))^2} ?
The integral of sqrt(1+(4/(\sqrt[3]{x)))^2} is (x^{2/3}+16)^{3/2}+C