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

The integral of (x^4+1)/(x^2) is (x^3)/3-1/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)

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

\frac{x ^{3}}{3} - \frac{1}{x} + C

Solution steps

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

Expand

\frac{x ^{4} + 1}{x ^{2}} : x^{2} + \frac{1}{x ^{2}}

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

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

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

= \frac{x ^{3}}{3} - \frac{1}{x}

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{1}{x} + C

\int x cos (\frac{π x}{4}) d x

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

\int ∣ 2 x - 5 ∣ d x

\int - x^{2} + 6 x - 8 d x

\int 24 (1 + x^{2}) (7 + 12 x + 4 x^{3})^{16} d x

What is the integral of (x^4+1)/(x^2) ?
The integral of (x^4+1)/(x^2) is (x^3)/3-1/x+C