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

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

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

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

Solution steps

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

Long division

\frac{x ^{4} + x ^{2} + 1}{x - 1} : x^{3} + x^{2} + 2 x + 2 + \frac{3}{x - 1}

= \int x^{3} + x^{2} + 2 x + 2 + \frac{3}{x - 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^{3} d x + \int x^{2} d x + \int 2 x d x + \int 2 d x + \int \frac{3}{x - 1} d x

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

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

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

\int 2 d x = 2 x

\int \frac{3}{x - 1} d x = 3 ln ∣ x - 1 ∣

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

Add a constant to the solution

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

\frac{d}{d x} (sin (x) - 3 cos (x))

\int \frac{12}{1 + t ^{2}} d t

- 5 sin (4 t)

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

\frac{s ^{3} + 5 s ^{2} + 9 s + 7}{s ^{2} + 3 s + 2}

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