Solve ∫ x^4+x^2+1/x^2+x+1 wrt x

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

Evaluate

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

Solution Steps

Factor the expressions that are not already factored in \frac{x^{4}+x^{2}+1}{x^{2}+x+1}.

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

Cancel out x^{2}+x+1 in both numerator and denominator.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}. Multiply -1 times \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.

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

Simplify.

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

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

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