What is the integral of 1/(x^{0.99)} ?

The integral of 1/(x^{0.99)} is 100x^{0.01}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^{0.99)}

\int \frac{1}{x ^{0.99}} d x

100 x^{0.01} + C

Solution steps

\int \frac{1}{x ^{0.99}} d x

Apply the Power Rule

: 100 x^{0.01}

= 100 x^{0.01}

Add a constant to the solution

= 100 x^{0.01} + C

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

\int (x - 2) (4 x + 2) d x

\int (x^{2} - x) (4 x^{3} - 6 x^{2} + 3)^{6} d x

\int 4.4 e^{(- 3 x + 4)} d x

\int \frac{2 x ^{3} - 4 x - 8}{x ( x - 1 ) ( x ^{2} + 4 )} d x

What is the integral of 1/(x^{0.99)} ?
The integral of 1/(x^{0.99)} is 100x^{0.01}+C