Solve ∫ 3/x^6-1/x^6+x wrt x

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

Evaluate

$$\frac{x^{2}}{2}-\frac{2}{5x^{5}}+С$$

Solution Steps

Integrate the sum term by term.

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

Factor out the constant in each of the terms.

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

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 3 times -\frac{1}{5x^{5}}.

$$-\frac{\frac{3}{x^{5}}}{5}-\int \frac{1}{x^{6}}\mathrm{d}x+\int x\mathrm{d}x$$

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}}.

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

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

$$-\frac{\frac{3}{x^{5}}}{5}+\frac{1}{5x^{5}}+\frac{x^{2}}{2}$$

Simplify.

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

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.

$$-\frac{2}{5x^{5}}+\frac{x^{2}}{2}+С$$

Differentiate w.r.t. x

$$x+\frac{2}{x^{6}}$$