Solve ∫ (from 0 to 1) of x^5-1/x-1 wrt x

$$\int _ { 0 } ^ { 1 } \frac { x ^ { 5 } - 1 } { x - 1 } d x$$

Evaluate

$$\frac{137}{60}\approx 2.283333333$$

Solution Steps

Factor the expressions that are not already factored in \frac{x^{5}-1}{x-1}.

$$\int _{0}^{1}\frac{\left(x-1\right)\left(x^{4}+x^{3}+x^{2}+x+1\right)}{x-1}\mathrm{d}x$$

Cancel out x-1 in both numerator and denominator.

$$\int _{0}^{1}x^{4}+x^{3}+x^{2}+x+1\mathrm{d}x$$

Evaluate the indefinite integral first.

$$\int x^{4}+x^{3}+x^{2}+x+1\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{4}\mathrm{d}x+\int x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}.

$$\frac{x^{5}}{5}+\int x^{3}\mathrm{d}x+\int x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 1\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

$$\frac{x^{5}}{5}+\frac{x^{4}}{4}+\int x^{2}\mathrm{d}x+\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}+\frac{x^{4}}{4}+\frac{x^{3}}{3}+\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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}.

$$\frac{x^{5}}{5}+\frac{x^{4}}{4}+\frac{x^{3}}{3}+\frac{x^{2}}{2}+\int 1\mathrm{d}x$$

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

$$\frac{x^{5}}{5}+\frac{x^{4}}{4}+\frac{x^{3}}{3}+\frac{x^{2}}{2}+x$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$1+\frac{1^{2}}{2}+\frac{1^{3}}{3}+\frac{1^{4}}{4}+\frac{1^{5}}{5}-\left(0+\frac{0^{2}}{2}+\frac{0^{3}}{3}+\frac{0^{4}}{4}+\frac{0^{5}}{5}\right)$$

Simplify.

$$\frac{137}{60}$$