What is the integral of (5-2x^2)/(y^2) ?

The integral of (5-2x^2)/(y^2) is -(5-2x^2)/y+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5-2x^2)/(y^2)

\int \frac{5 - 2 x ^{2}}{y ^{2}} d y

- \frac{5 - 2 x ^{2}}{y} + C

Solution steps

\int \frac{5 - 2 x ^{2}}{y ^{2}} d y

Take the constant out:

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

= (5 - 2 x^{2}) \cdot \int \frac{1}{y ^{2}} d y

\int \frac{1}{y ^{2}} d y = - \frac{1}{y}

= (5 - 2 x^{2}) (- \frac{1}{y})

Simplify

(5 - 2 x^{2}) (- \frac{1}{y}) : - \frac{5 - 2 x ^{2}}{y}

= - \frac{5 - 2 x ^{2}}{y}

Add a constant to the solution

= - \frac{5 - 2 x ^{2}}{y} + C

\int sin (10 x) d x

y = x (x + 3) (x - 4), x = 0

(5 e^{x})^{^{'}}

x \to 0 lim (\frac{1}{tan ( x )})

\int_{b}^{2 b} x^{8} d x

What is the integral of (5-2x^2)/(y^2) ?
The integral of (5-2x^2)/(y^2) is -(5-2x^2)/y+C