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{sin ( x )}{cos ^{2} ( x )}) d x

n = 2 \sum \infty \frac{8}{4 ^{2 n}}

y^{^{'}} = \frac{- 2 x y}{1 + x ^{2}}

\frac{\partial}{\partial x} (42 y^{8} x^{5} - 30 y^{7} x^{4})

cos (2 arctan (\frac{x}{2}))

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