Solve ∫ 3/t^3dt=

$$\int \frac { 3 } { t ^ { 3 } } d t =$$

Evaluate

$$-\frac{3}{2t^{2}}+С$$

Solution Steps

Factor out the constant using \int af\left(t\right)\mathrm{d}t=a\int f\left(t\right)\mathrm{d}t.

$$3\int \frac{1}{t^{3}}\mathrm{d}t$$

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

$$-\frac{\frac{3}{t^{2}}}{2}$$

Simplify.

$$-\frac{3}{2t^{2}}$$

If F\left(t\right) is an antiderivative of f\left(t\right), then the set of all antiderivatives of f\left(t\right) is given by F\left(t\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$-\frac{3}{2t^{2}}+С$$

Differentiate w.r.t. t

$$\frac{3}{t^{3}}$$