Solve ∫ x^3-1/x-1 wrt x

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

Evaluate

$$\frac{x^{3}}{3}+\frac{x^{2}}{2}+x+С$$

Solution Steps

Factor the expressions that are not already factored in \frac{x^{3}-1}{x-1}.

$$\int \frac{\left(x-1\right)\left(x^{2}+x+1\right)}{x-1}\mathrm{d}x$$

Cancel out x-1 in both numerator and denominator.

$$\int x^{2}+x+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}.

$$\frac{x^{3}}{3}+\int x\mathrm{d}x+\int 1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}.

$$\frac{x^{3}}{3}+\frac{x^{2}}{2}+\int 1\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+С$$

Differentiate w.r.t. x

$$x^{2}+x+1$$