Solve ∫ 1/mu^2du=4p/2

$$\int \frac { 1 } { \mu ^ { 2 } } d u = \frac { 4 p } { 2 }$$

Solve for p

$$p=\frac{u}{2\mu ^{2}}+С$$

$$\mu \neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2.

$$2\int \frac{1}{\mu ^{2}}\mathrm{d}u=4p$$

Swap sides so that all variable terms are on the left hand side.

$$4p=2\int \frac{1}{\mu ^{2}}\mathrm{d}u$$

The equation is in standard form.

$$4p=2С+\frac{2u}{\mu ^{2}}$$

Divide both sides by 4.

$$\frac{4p}{4}=\frac{\frac{2u}{\mu ^{2}}+С}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$p=\frac{\frac{2u}{\mu ^{2}}+С}{4}$$

Divide С+\frac{2u}{\mu ^{2}} by 4.

$$p=\frac{u}{2\mu ^{2}}+С$$

Solve for μ

$$\left\{\begin{matrix}\mu =\sqrt{-\frac{u}{С-2p}}\text{; }\mu =-\sqrt{-\frac{u}{С-2p}}\text{, }&\left(С\leq 2p\text{ and }p\neq \frac{С_{1}}{2}\text{ and }u>0\text{ and }p\geq С_{2}\right)\text{ or }\left(С_{3}\geq 2p\text{ and }p\neq \frac{С_{4}}{2}\text{ and }u<0\text{ and }p\leq С_{5}\right)\\\mu \neq 0\text{, }&p=\frac{С}{2}\text{ and }u=0\end{matrix}\right.$$