Solve ∫ (3x^2-2x+1/sqrt{x}) wrt x

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

Evaluate

$$x^{3}-x^{2}+2\sqrt{x}+С$$

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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

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

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

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

Rewrite \frac{1}{\sqrt{x}} as x^{-\frac{1}{2}}. Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{-\frac{1}{2}}\mathrm{d}x with \frac{x^{\frac{1}{2}}}{\frac{1}{2}}. Simplify and convert from exponential to radical form.

$$x^{3}-x^{2}+2\sqrt{x}$$

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.

$$x^{3}-x^{2}+2\sqrt{x}+С$$

Differentiate w.r.t. x

$$3x^{2}-2x+\frac{1}{\sqrt{x}}$$