What is the integral of 1/(\sqrt[3]{9x-1)} ?

The integral of 1/(\sqrt[3]{9x-1)} is 1/6 (9x-1)^{2/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(\sqrt[3]{9x-1)}

\int \frac{1}{3 9 x - 1} d x

\frac{1}{6} (9 x - 1)^{\frac{2}{3}} + C

Solution steps

\int \frac{1}{3 9 x - 1} d x

Apply u-substitution

: \int \frac{1}{9 3 u} d u

= \int \frac{1}{9 3 u} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{3 u} d u

Apply the Power Rule

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

= \frac{1}{9} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 9 x - 1

= \frac{1}{9} \cdot \frac{3}{2} (9 x - 1)^{\frac{2}{3}}

Simplify

\frac{1}{9} \cdot \frac{3}{2} (9 x - 1)^{\frac{2}{3}} : \frac{1}{6} (9 x - 1)^{\frac{2}{3}}

= \frac{1}{6} (9 x - 1)^{\frac{2}{3}}

Add a constant to the solution

= \frac{1}{6} (9 x - 1)^{\frac{2}{3}} + C

\int 81 - x^{2} d x

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

f (x) = x^{2} - x + 1

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

x \to 0 lim (1 - cos (x))

What is the integral of 1/(\sqrt[3]{9x-1)} ?
The integral of 1/(\sqrt[3]{9x-1)} is 1/6 (9x-1)^{2/3}+C