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

\int \frac{1}{3 x ln ( x )} d x

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

\int (\frac{1}{5} - \frac{2}{x ^{3}} + 2 x) d x

\int x^{3} - 2 d x

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