Solve ∫ 0.2/sqrtx^3

$$\int \frac { 0.2 } { \sqrt { x } ^ { 3 } }$$

Evaluate

$$-\frac{2}{5\sqrt{x}}+С$$

Solution Steps

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

$$\frac{\int \frac{1}{x^{\frac{3}{2}}}\mathrm{d}x}{5}$$

Rewrite \frac{1}{x^{\frac{3}{2}}} as x^{-\frac{3}{2}}. 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{x^{-\frac{1}{2}}}{-\frac{1}{2}}. Simplify and convert from exponential to radical form.

$$-\frac{\frac{2}{\sqrt{x}}}{5}$$

Simplify.

$$-\frac{2}{5\sqrt{x}}$$

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.

$$-\frac{2}{5\sqrt{x}}+С$$

Differentiate w.r.t. x

$$\frac{1}{5x^{1.5}}$$