What is the integral of ((x^4-5x^2+3x-4))/x ?

The integral of ((x^4-5x^2+3x-4))/x is (x^4)/4-(5x^2)/2+3x-4ln|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^4-5x^2+3x-4))/x

\int \frac{( x ^{4} - 5 x ^{2} + 3 x - 4 )}{x} d x

\frac{x ^{4}}{4} - \frac{5 x ^{2}}{2} + 3 x - 4 ln ∣ x ∣ + C

Solution steps

\int \frac{x ^{4} - 5 x ^{2} + 3 x - 4}{x} d x

Expand

\frac{x ^{4} - 5 x ^{2} + 3 x - 4}{x} : x^{3} - 5 x + 3 - \frac{4}{x}

= \int x^{3} - 5 x + 3 - \frac{4}{x} 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^{3} d x - \int 5 x d x + \int 3 d x - \int \frac{4}{x} d x

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

\int 5 x d x = \frac{5 x ^{2}}{2}

\int 3 d x = 3 x

\int \frac{4}{x} d x = 4 ln ∣ x ∣

= \frac{x ^{4}}{4} - \frac{5 x ^{2}}{2} + 3 x - 4 ln ∣ x ∣

Add a constant to the solution

= \frac{x ^{4}}{4} - \frac{5 x ^{2}}{2} + 3 x - 4 ln ∣ x ∣ + C

\int_{1}^{e} (\frac{x ^{2} - 1}{x}) d x

\frac{d}{d x} (ln (\frac{x + y}{x - y}))

\frac{\partial}{\partial s} (sin (s^{2} - t^{2}))

\frac{d}{d x} (ln (2 x^{4} e^{3 x} (x^{2} + x + 1)^{5}))

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

What is the integral of ((x^4-5x^2+3x-4))/x ?
The integral of ((x^4-5x^2+3x-4))/x is (x^4)/4-(5x^2)/2+3x-4ln|x|+C