Solve ∫ x^2-2x^2+4x/x

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

Evaluate

$$-\frac{x^{2}}{2}+4x+С$$

Solution Steps

Combine x^{2} and -2x^{2} to get -x^{2}.

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

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

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

Cancel out x in both numerator and denominator.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

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

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.

$$-\frac{x^{2}}{2}+4x+С$$

Differentiate w.r.t. x

$$4-x$$