What is the integral of (9/(x^4)+x^2) ?

The integral of (9/(x^4)+x^2) is -3/(x^3)+(x^3)/3+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (9/(x^4)+x^2)

\int (\frac{9}{x ^{4}} + x^{2}) d x

- \frac{3}{x ^{3}} + \frac{x ^{3}}{3} + C

Solution steps

\int \frac{9}{x ^{4}} + x^{2} 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{9}{x ^{4}} d x + \int x^{2} d x

\int \frac{9}{x ^{4}} d x = - \frac{3}{x ^{3}}

\int x^{2} d x = \frac{x ^{3}}{3}

= - \frac{3}{x ^{3}} + \frac{x ^{3}}{3}

Add a constant to the solution

= - \frac{3}{x ^{3}} + \frac{x ^{3}}{3} + C

What is the integral of (9/(x^4)+x^2) ?
The integral of (9/(x^4)+x^2) is -3/(x^3)+(x^3)/3+C