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

The integral of sqrt(1+(2/3 x^{-1/3))^2} is 1/27 (9x^{2/3}+4)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{1}{27} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}} + C

Solution steps

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

Simplify

1 + (\frac{2}{3} x^{- \frac{1}{3}})^{2} : \frac{( 9 x ^{\frac{2}{3}} + 4 ) ^{\frac{1}{2}}}{3 ( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}}

= \int \frac{( 9 x ^{\frac{2}{3}} + 4 ) ^{\frac{1}{2}}}{3 ( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{( 9 x ^{\frac{2}{3}} + 4 ) ^{\frac{1}{2}}}{( x ^{\frac{2}{3}} ) ^{\frac{1}{2}}} d x

Apply u-substitution

= \frac{1}{3} \cdot \int \frac{u ^{\frac{1}{2}}}{6} d u

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{6} \cdot \int u^{\frac{1}{2}} d u

Apply the Power Rule

= \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 9 x^{\frac{2}{3}} + 4

= \frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}}

Simplify

\frac{1}{3} \cdot \frac{1}{6} \cdot \frac{2}{3} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}} : \frac{1}{27} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}}

= \frac{1}{27} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{27} (9 x^{\frac{2}{3}} + 4)^{\frac{3}{2}} + C

What is the integral of sqrt(1+(2/3 x^{-1/3))^2} ?
The integral of sqrt(1+(2/3 x^{-1/3))^2} is 1/27 (9x^{2/3}+4)^{3/2}+C