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

\frac{d}{d x} (sec (\frac{π}{6}))

\int x^{2} (x^{3} + 8)^{3} d x

x^{2} y^{^{'}} + x (x + 2) y = e^{x}

x = 2 y, x = y^{2} - 8

x^{a} (1 - x)^{b}

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