What is the integral of 1/(z^2sqrt(a^2-x^2)) ?

The integral of 1/(z^2sqrt(a^2-x^2)) is -1/(zsqrt(a^2-x^2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(z^2sqrt(a^2-x^2))

\int \frac{1}{z ^{2} a ^{2} - x ^{2}} d z

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

Solution steps

\int \frac{1}{z ^{2} a ^{2} - x ^{2}} d z

Take the constant out:

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

= \frac{1}{a ^{2} - x ^{2}} \cdot \int \frac{1}{z ^{2}} d z

Apply the Power Rule

= \frac{1}{a ^{2} - x ^{2}} (- \frac{1}{z})

Simplify

\frac{1}{a ^{2} - x ^{2}} (- \frac{1}{z}) : - \frac{1}{z a ^{2} - x ^{2}}

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

Add a constant to the solution

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

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