Solve ∫ [2x^4+2x^3-8x^2-8x/x+1] wrt x

$$\int [ \frac { 2 x ^ { 4 } + 2 x ^ { 3 } - 8 x ^ { 2 } - 8 x } { x + 1 } ] d x$$

Evaluate

$$\frac{x^{4}}{2}-4x^{2}+С$$

Solution Steps

Factor the expressions that are not already factored in \frac{2x^{4}+2x^{3}-8x^{2}-8x}{x+1}.

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

Cancel out x+1 in both numerator and denominator.

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

Expand the expression.

$$\int 2x^{3}-8x\mathrm{d}x$$

Integrate the sum term by term.

$$\int 2x^{3}\mathrm{d}x+\int -8x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$2\int x^{3}\mathrm{d}x-8\int x\mathrm{d}x$$

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}. Multiply 2 times \frac{x^{4}}{4}.

$$\frac{x^{4}}{2}-8\int x\mathrm{d}x$$

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 -8 times \frac{x^{2}}{2}.

$$\frac{x^{4}}{2}-4x^{2}$$

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.

$$-4x^{2}+\frac{x^{4}}{2}+С$$

Differentiate w.r.t. x

$$2x\left(x^{2}-4\right)$$