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

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

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

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

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

Solution steps

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

Simplify

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

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

Apply u-substitution

: \int \frac{3 u ^{\frac{1}{2}}}{2} d u

= \int \frac{3 u ^{\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^{\frac{1}{2}} d u

Apply the Power Rule

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

= (x^{\frac{2}{3}} + 1)^{\frac{3}{2}} + C

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y = x 6^{x}

x (x + y) (\frac{d y}{d x}) + y (3 x + y) = 0

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\int_{1}^{3} 5 r^{3} ln (r) d r

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