Solve ∫ 6sqrt{x}

$$\int 6 \sqrt { x }$$

Evaluate

$$4x^{\frac{3}{2}}+С$$

Solution Steps

Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.

$$6\int \sqrt{x}\mathrm{d}x$$

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. Multiply 6 times \frac{2x^{\frac{3}{2}}}{3}.

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

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.

$$4x^{\frac{3}{2}}+С$$

Differentiate w.r.t. x

$$6\sqrt{x}$$