What is the integral of (2z^2)/y ?

The integral of (2z^2)/y is 2z^2ln|y|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2z^2)/y

\int \frac{2 z ^{2}}{y} d y

2 z^{2} ln ∣ y ∣ + C

Solution steps

\int \frac{2 z ^{2}}{y} d y

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{y} d y = ln (∣ y ∣)

= 2 z^{2} ln ∣ y ∣

Add a constant to the solution

= 2 z^{2} ln ∣ y ∣ + C

\int \frac{ln ( x ) cos ( ln ( x ) )}{x} d x

\int \frac{2 x ^{2} + 5 x + 16}{x ^{4} + 8 x ^{2} + 16} d x

\int \frac{ln ( 2 x )}{ln ( 4 x )} \frac{1}{x} d x

\int sin^{2} (\frac{x}{8}) cos (\frac{x}{8}) d x

\int \frac{x}{49 - x ^{2}} d x