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

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

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

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

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

Solution steps

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

Long division

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

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

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

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

\int 2 d x = 2 x

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

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

Add a constant to the solution

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

h \to 0 lim (\frac{1}{h})

y = (3 - x^{2}) (x^{3} - 3 x + 2)

\int_{5}^{8} \frac{1}{x ^{2} - 1} d x

\frac{d y}{d x} = x y + 5 x

x \to 0 lim (\frac{x}{sin ( 5 x )})

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