حل ∫ 1/x^3 wrt x=

$$\int \frac { 1 } { x ^ { 3 } } d x =$$

تقدير القيمة

$$-\frac{1}{2x^{2}}+С$$

خطوات الحل

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}}.

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

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.

$$-\frac{1}{2x^{2}}+С$$

تفاضل w.r.t. x

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