What is the integral of sqrt(1+(16)/(x^{2/3))} ?

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

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integral of sqrt(1+(16)/(x^{2/3))}

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

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

Solution steps

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

Apply u-substitution

: \int \frac{3 ( u + 16 ) ^{\frac{1}{2}}}{2} d u

= \int \frac{3 ( u + 16 ) ^{\frac{1}{2}}}{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)^{\frac{1}{2}} d u

Apply u-substitution

: \int v^{\frac{1}{2}} d v

= \frac{3}{2} \cdot \int v^{\frac{1}{2}} 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

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