Solve ∫ 3x^2+2x-8/x+2 wrt x

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

Evaluate

$$\frac{3x^{2}}{2}-4x+С$$

Solution Steps

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

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

Cancel out x+2 in both numerator and denominator.

$$\int 3x-4\mathrm{d}x$$

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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 3 times \frac{x^{2}}{2}.

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

Find the integral of -4 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

Simplify.

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

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.

$$-4x+\frac{3x^{2}}{2}+С$$

Differentiate w.r.t. x

$$3x-4$$