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

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

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

integral

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

2 x + \frac{1}{3} (x - 1)^{3} + \frac{3}{2} (x - 1)^{2} - \frac{1}{2} (x - 1)^{2} - 2 + C

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

\frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣ - (\frac{( x - 1 ) ^{2}}{2} + 2 (x - 1) + ln ∣ x - 1 ∣) + x - 1 + ln ∣ x - 1 ∣ - ln ∣ x - 1 ∣ : 2 x + \frac{1}{3} (x - 1)^{3} + \frac{3}{2} (x - 1)^{2} - \frac{1}{2} (x - 1)^{2} - 2

= 2 x + \frac{1}{3} (x - 1)^{3} + \frac{3}{2} (x - 1)^{2} - \frac{1}{2} (x - 1)^{2} - 2

Add a constant to the solution

= 2 x + \frac{1}{3} (x - 1)^{3} + \frac{3}{2} (x - 1)^{2} - \frac{1}{2} (x - 1)^{2} - 2 + C

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

\frac{5}{x ^{2}}

(2 x - 1) e^{x}

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

\frac{5 x ^{2}}{( x - 2 ) ( x ^{2} + 1 )}

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