Solve ∫ (sqrt{x}-1) wrt x=

$$\int ( \sqrt { x } - 1 ) d x =$$

Evaluate

$$\frac{2x^{\frac{3}{2}}}{3}-x+С$$

Solution Steps

Integrate the sum term by term.

$$\int \sqrt{x}\mathrm{d}x+\int -1\mathrm{d}x$$

Rewrite \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{3}{2}}}{\frac{3}{2}}. Simplify.

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

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

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

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.

$$\frac{2x^{\frac{3}{2}}}{3}-x+С$$

Differentiate w.r.t. x

$$\sqrt{x}-1$$