What is the integral of 1/(sqrt(x^6)) ?

The integral of 1/(sqrt(x^6)) is -1/(2x^2)+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(x^6))

\int \frac{1}{x ^{6}} d x

- \frac{1}{2 x ^{2}} + C

Solution steps

\int \frac{1}{x ^{6}} d x

x^{6} = x^{3},

assuming

x \geq 0

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

Apply the Power Rule

: - \frac{1}{2 x ^{2}}

= - \frac{1}{2 x ^{2}}

Add a constant to the solution

= - \frac{1}{2 x ^{2}} + C

\int e^{- 2 b x^{2}} d x

\int \frac{9}{x x ^{2} - 1} d x

\int 3 cos^{2} (2 x) d x

\int \frac{4}{5} x^{\frac{4}{5}} d x

\int 9 - y^{2} d y

What is the integral of 1/(sqrt(x^6)) ?
The integral of 1/(sqrt(x^6)) is -1/(2x^2)+C