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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

x \to 0 lim (e^{x^{2}})

\int_{1}^{2} x^{3} ln (x) d x

\int 2 sin (x) + 1 d x

\int_{0}^{\frac{π}{3}} (sec^{2} (x)) d x

y = sin (x cos (x))

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