What is the integral of ((x^5+1))/((x-1)) ?

The integral of ((x^5+1))/((x-1)) is (x^5)/5+(x^4)/4+(x^3)/3+(x^2)/2+x+2ln|x-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((x^5+1))/((x-1))

integral

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

\frac{x ^{5}}{5} + \frac{x ^{4}}{4} + \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + 2 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{x ^{5} + 1}{x - 1} d x

Expand

\frac{x ^{5} + 1}{x - 1} : \frac{x ^{5}}{x - 1} + \frac{1}{x - 1}

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{x ^{5}}{x - 1} d x + \int \frac{1}{x - 1} d x

\int \frac{x ^{5}}{x - 1} d x = \frac{x ^{5}}{5} + \frac{x ^{4}}{4} + \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + ln ∣ x - 1 ∣

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

= \frac{x ^{5}}{5} + \frac{x ^{4}}{4} + \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + ln ∣ x - 1 ∣ + ln ∣ x - 1 ∣

Simplify

= \frac{x ^{5}}{5} + \frac{x ^{4}}{4} + \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + 2 ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{x ^{5}}{5} + \frac{x ^{4}}{4} + \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + 2 ln ∣ x - 1 ∣ + C

\frac{e ^{x^{2}}}{x}

\frac{x ^{2} - 12 x + 13}{x}

\frac{( x + 3 )}{( x - 1 )}

\frac{1}{( e ^{x^{2}} )}

\frac{( 2 x + 1 )}{( 3 x + 2 )}

What is the integral of ((x^5+1))/((x-1)) ?
The integral of ((x^5+1))/((x-1)) is (x^5)/5+(x^4)/4+(x^3)/3+(x^2)/2+x+2ln|x-1|+C