Solve ∫ sqrt{x}sqrt{x}sqrt{x wrt x

$$\int \sqrt { x } \sqrt { x } \sqrt { x } d x$$

Evaluate

$$\frac{2x^{\frac{5}{2}}}{5}+С$$

Solution Steps

Multiply \sqrt{x} and \sqrt{x} to get \left(\sqrt{x}\right)^{2}.

$$\int \left(\sqrt{x}\right)^{2}\sqrt{x}\mathrm{d}x$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$\int \left(\sqrt{x}\right)^{3}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{\frac{3}{2}}\mathrm{d}x with \frac{2x^{\frac{5}{2}}}{5}.

$$\frac{2x^{\frac{5}{2}}}{5}$$

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.

$$\frac{2x^{\frac{5}{2}}}{5}+С$$

Differentiate w.r.t. x

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