What is the integral of x^4-1/(x^2)+1 ?

The integral of x^4-1/(x^2)+1 is (x^5)/5+1/x+x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^4-1/(x^2)+1

\int x^{4} - \frac{1}{x ^{2}} + 1 d x

\frac{x ^{5}}{5} + \frac{1}{x} + x + C

Solution steps

\int x^{4} - \frac{1}{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

= \int x^{4} d x - \int \frac{1}{x ^{2}} d x + \int 1 d x

\int x^{4} d x = \frac{x ^{5}}{5}

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

\int 1 d x = x

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

Simplify

= \frac{x ^{5}}{5} + \frac{1}{x} + x

Add a constant to the solution

= \frac{x ^{5}}{5} + \frac{1}{x} + x + C

\int sin (a x) - e^{\frac{x}{b}} d x

\int x (x + 5) d x

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

\int (\frac{5}{1 - x ^{2}} + \frac{3}{x}) d x

What is the integral of x^4-1/(x^2)+1 ?
The integral of x^4-1/(x^2)+1 is (x^5)/5+1/x+x+C