What is the integral of 1/z 1/x (x+z/x) ?

The integral of 1/z 1/x (x+z/x) is -1/x+x/z+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/z 1/x (x+z/x)

\int \frac{1}{z} \frac{1}{x} (x + \frac{z}{x}) d x

- \frac{1}{x} + \frac{x}{z} + C

Solution steps

\int \frac{1}{z} \cdot \frac{1}{x} (x + \frac{z}{x}) d x

Take the constant out:

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

= \frac{1}{z} \cdot \int (\frac{z}{x} + x) \frac{1}{x} d x

Simplify

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

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

Expand

\frac{z + x ^{2}}{x ^{2}} : \frac{z}{x ^{2}} + 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{z} (\int \frac{z}{x ^{2}} d x + \int 1 d x)

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

\int 1 d x = x

= \frac{1}{z} (- \frac{z}{x} + x)

Simplify

\frac{1}{z} (- \frac{z}{x} + x) : - \frac{1}{x} + \frac{x}{z}

= - \frac{1}{x} + \frac{x}{z}

Add a constant to the solution

= - \frac{1}{x} + \frac{x}{z} + C

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

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

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

\int (x^{2} sin (2 x)) d x

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

What is the integral of 1/z 1/x (x+z/x) ?
The integral of 1/z 1/x (x+z/x) is -1/x+x/z+C