Solve ∫ 6x^2+-4/x^3 wrt x

$$\int 6 x ^ { 2 } + \frac { - 4 } { x ^ { 3 } } d x$$

Evaluate

$$2x^{3}+\frac{2}{x^{2}}+С$$

Solution Steps

Integrate the sum term by term.

$$\int 6x^{2}\mathrm{d}x+\int -\frac{4}{x^{3}}\mathrm{d}x$$

Factor out the constant in each of the terms.

$$6\int x^{2}\mathrm{d}x-4\int \frac{1}{x^{3}}\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 6 times \frac{x^{3}}{3}.

$$2x^{3}-4\int \frac{1}{x^{3}}\mathrm{d}x$$

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}}.

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

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.

$$2x^{3}+\frac{2}{x^{2}}+С$$

Differentiate w.r.t. x

$$6x^{2}-\frac{4}{x^{3}}$$