Solve ∫ x^2(1-1/x) wrt x

$$\int x ^ { 2 } ( 1 - \frac { 1 } { x } ) d x$$

Evaluate

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}+С$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

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

Since \frac{x}{x} and \frac{1}{x} have the same denominator, subtract them by subtracting their numerators.

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

Express x^{2}\times \frac{x-1}{x} as a single fraction.

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

Cancel out x in both numerator and denominator.

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

Use the distributive property to multiply x by x-1.

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

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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}.

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

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}.

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

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.

$$\frac{x^{3}}{3}-\frac{x^{2}}{2}+С$$

Differentiate w.r.t. x

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