Solve ∫ 2x^4+3x^3-2x^2-x+3/2x+3 wrt x

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

Evaluate

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

Solution Steps

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

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

Cancel out 2x+3 in both numerator and denominator.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}.

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

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

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

Simplify.

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

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.

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

Differentiate w.r.t. x

$$x^{3}-x+1$$