What is the integral of (2x)/y-(3y^2)/(x^4) ?

The integral of (2x)/y-(3y^2)/(x^4) is (x^2)/y+(y^2)/(x^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (2x)/y-(3y^2)/(x^4)

\int \frac{2 x}{y} - \frac{3 y ^{2}}{x ^{4}} d x

\frac{x ^{2}}{y} + \frac{y ^{2}}{x ^{3}} + C

Solution steps

\int \frac{2 x}{y} - \frac{3 y ^{2}}{x ^{4}} 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

= \int \frac{2 x}{y} d x - \int \frac{3 y ^{2}}{x ^{4}} d x

\int \frac{2 x}{y} d x = \frac{x ^{2}}{y}

\int \frac{3 y ^{2}}{x ^{4}} d x = - \frac{y ^{2}}{x ^{3}}

= \frac{x ^{2}}{y} - (- \frac{y ^{2}}{x ^{3}})

Simplify

= \frac{x ^{2}}{y} + \frac{y ^{2}}{x ^{3}}

Add a constant to the solution

= \frac{x ^{2}}{y} + \frac{y ^{2}}{x ^{3}} + C

What is the integral of (2x)/y-(3y^2)/(x^4) ?
The integral of (2x)/y-(3y^2)/(x^4) is (x^2)/y+(y^2)/(x^3)+C