Solve ∫ 1/4+4/x^3 wrt x

$$\int{ \frac{ 1 }{ 4 } + \frac{ 4 }{ { x }^{ 3 } } }d x$$

Evaluate

$$\frac{x}{4}-\frac{2}{x^{2}}+С$$

Solution Steps

Integrate the sum term by term.

$$\int \frac{1}{4}\mathrm{d}x+\int \frac{4}{x^{3}}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int \frac{1}{4}\mathrm{d}x+4\int \frac{1}{x^{3}}\mathrm{d}x$$

Find the integral of \frac{1}{4} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x}{4}+4\int \frac{1}{x^{3}}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{x^{3}}\mathrm{d}x with -\frac{1}{2x^{2}}. Multiply 4 times -\frac{1}{2x^{2}}.

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

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.

$$\frac{x}{4}-\frac{2}{x^{2}}+С$$

Differentiate w.r.t. x

$$\frac{1}{4}+\frac{4}{x^{3}}$$