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

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

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

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

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + C

Solution steps

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

Simplify

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

= \int x^{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 1 d x

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

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

\int 1 d x = x

= \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{x ^{2}}{2} + x + C

\int_{\frac{π}{2}}^{π} x sin^{2} (x) d x

\frac{d}{d x} (\frac{1 - y}{x - 1})

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

\int sin^{2} (t) cos^{3} (t) d t

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

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