What is the integral of (7x^3-4x+2)/(x^3-x^2) ?

The integral of (7x^3-4x+2)/(x^3-x^2) is (7x^2)/(x-1)+5ln|x-1|-7/(x-1)+2ln|x|+2/x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (7x^3-4x+2)/(x^3-x^2)

\int \frac{7 x ^{3} - 4 x + 2}{x ^{3} - x ^{2}} d x

\frac{7 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{7}{x - 1} + 2 ln ∣ x ∣ + \frac{2}{x} + C

Solution steps

\int \frac{7 x ^{3} - 4 x + 2}{x ^{3} - x ^{2}} d x

Expand

\frac{7 x ^{3} - 4 x + 2}{x ^{3} - x ^{2}} : \frac{7 x ^{3}}{x ^{3} - x ^{2}} - \frac{4 x}{x ^{3} - x ^{2}} + \frac{2}{x ^{3} - x ^{2}}

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

\int \frac{7 x ^{3}}{x ^{3} - x ^{2}} d x = 7 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1})

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

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

= 7 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - 4 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) + 2 (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣)

Simplify

7 (\frac{x ^{2}}{x - 1} + ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - 4 (- ln ∣ x ∣ + ln ∣ x - 1 ∣) + 2 (- ln ∣ x ∣ + \frac{1}{x} + ln ∣ x - 1 ∣) : \frac{7 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{7}{x - 1} + 2 ln ∣ x ∣ + \frac{2}{x}

= \frac{7 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{7}{x - 1} + 2 ln ∣ x ∣ + \frac{2}{x}

Add a constant to the solution

= \frac{7 x ^{2}}{x - 1} + 5 ln ∣ x - 1 ∣ - \frac{7}{x - 1} + 2 ln ∣ x ∣ + \frac{2}{x} + C

\int sin^{2} (x) d x

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

\int \frac{tan ( y )}{cos ^{2} ( y )} d y

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

\int \frac{2 x + 9}{x ^{2} + 9 x + 2} d x

What is the integral of (7x^3-4x+2)/(x^3-x^2) ?
The integral of (7x^3-4x+2)/(x^3-x^2) is (7x^2)/(x-1)+5ln|x-1|-7/(x-1)+2ln|x|+2/x+C