Solve ∫ 1/x^4-1/x^6 wrt x

$$\int{ \frac{ 1 }{ { x }^{ 4 } } - \frac{ 1 }{ { x }^{ 6 } } }d x$$

Evaluate

$$-\frac{1}{3x^{3}}+\frac{1}{5x^{5}}+С$$

Solution Steps

Integrate the sum term by term.

$$\int \frac{1}{x^{4}}\mathrm{d}x+\int -\frac{1}{x^{6}}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int \frac{1}{x^{4}}\mathrm{d}x-\int \frac{1}{x^{6}}\mathrm{d}x$$

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

$$-\frac{1}{3x^{3}}-\int \frac{1}{x^{6}}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{x^{6}}\mathrm{d}x with -\frac{1}{5x^{5}}. Multiply -1 times -\frac{1}{5x^{5}}.

$$-\frac{1}{3x^{3}}+\frac{1}{5x^{5}}$$

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.

$$-\frac{1}{3x^{3}}+\frac{1}{5x^{5}}+С$$

Differentiate w.r.t. x

$$\frac{x^{2}-1}{x^{6}}$$