Solve ∫ 6x^4+4x^3-2x^2/2x wrt x

$$\int \frac { 6 x ^ { 4 } + 4 x ^ { 3 } - 2 x ^ { 2 } } { 2 x } d x$$

Evaluate

$$\frac{3x^{4}}{4}+\frac{2x^{3}}{3}-\frac{x^{2}}{2}+С$$

Solution Steps

Factor the expressions that are not already factored in \frac{6x^{4}+4x^{3}-2x^{2}}{2x}.

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

Cancel out 2x in both numerator and denominator.

$$\int x\left(3x-1\right)\left(x+1\right)\mathrm{d}x$$

Expand the expression.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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 3 times \frac{x^{4}}{4}.

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

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

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

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 -1 times \frac{x^{2}}{2}.

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

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.

$$\frac{3x^{4}}{4}+\frac{2x^{3}}{3}-\frac{x^{2}}{2}+С$$

Differentiate w.r.t. x

$$x\left(3x-1\right)\left(x+1\right)$$