Solve ∫ 4x^4-2x^3+x^2/x^2 wrt x=

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

Evaluate

$$\frac{4x^{3}}{3}-x^{2}+x+С$$

Solution Steps

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

$$\int \frac{x^{2}\left(4x^{2}-2x+1\right)}{x^{2}}\mathrm{d}x$$

Cancel out x^{2} in both numerator and denominator.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 4 times \frac{x^{3}}{3}.

$$\frac{4x^{3}}{3}-2\int x\mathrm{d}x+\int 1\mathrm{d}x$$

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

$$\frac{4x^{3}}{3}-x^{2}+\int 1\mathrm{d}x$$

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

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

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.

$$\frac{4x^{3}}{3}-x^{2}+x+С$$

Differentiate w.r.t. x

$$4x^{2}-2x+1$$