You use the integrating factor when the differential equation is not exact.
1. First Order Linear Ordinary Differential Equations
If the differential equation is in the form $\frac{\mathrm{d} y}{\mathrm{d} x}+f(x)\cdot y=g(x)$, you could use the integrating factor, $\mu(x)$. Keep this form in mind.
Remember the product rule? It’s $\frac{\mathrm{d} }{\mathrm{d} x}(f\cdot g)=f’\cdot g+f\cdot g’$.
Let’s replace $f$ with the $y$ and $g$ with a function that we don’t know $\mu(x)$, so that it satisfies $\frac{\mathrm{d} }{\mathrm{d} x}(\mu(x)\cdot y)=\mu’(x)\cdot y+\frac{\mathrm{d} y}{\mathrm{d} x}\cdot \mu(x)$, so that you can get a form that’s similar with $\frac{\mathrm{d} y}{\mathrm{d} x}+f(x)\cdot y=g(x)$.
If you multiply everything by the integrating factor, you get $\mu(x)\frac{\mathrm{d} y}{\mathrm{d} x}+\mu(x)\cdot f(x)\cdot y=\mu(x) \cdot g(x)$.
From above:
$\frac{\mathrm{d} }{\mathrm{d} x}(\mu(x)\cdot y)=\mu’(x)\cdot y+\frac{\mathrm{d} y}{\mathrm{d} x}\cdot \mu(x)$ $\mu(x)\frac{\mathrm{d} y}{\mathrm{d} x}+\mu(x)\cdot f(x)\cdot y=\mu(x) \cdot g(x)$
Comparing where the $y$’s are: $\mu(x)\cdot f(x)=\mu’(x)$
$\frac{\mu’(x)}{\mu(x)}=f(x)$
$\int \frac{\mu’(x)}{\mu(x)}\mathrm{d}x =\int f(x)\mathrm{d}x$
$\mathrm{ln}|\mu(x)|=\int f(x)\mathrm{d}x$
$\mu(x)=e^{\int f(x)\mathrm{d}x}$. Do not worry about plus-minus - you only need one integrating factor.
So, substitute this $\mu(x)$ into $\mu(x)\frac{\mathrm{d} y}{\mathrm{d} x}+\mu(x)\cdot f(x)\cdot y=\mu(x) \cdot g(x)$ and you notice: the left hand side is $\frac{\mathrm{d} }{\mathrm{d} x}(\mu(x)\cdot y)$
$\frac{\mathrm{d} }{\mathrm{d} x}(\mu(x)\cdot y)=\mu(x)\cdot g(x)$
You can then solve the equation from there.
For reference: $y \cdot e^{f(x)}=\int g(x)\cdot \mu(x)\mathrm{d}x +c$
1.1. Example
$\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{2}{5}y=3e^{-x}$ , $y(0)=5$
Recall that your integrating factor $\mu(x)$ is $e^{\int f(x) \mathrm{d}x}$, so in this case, $\mu(x)=e^{\int \frac{2}{5} \mathrm{d}x}$.
$e^{\frac{2}{5}x}\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{2}{5}e^{\frac{2}{5}x}y=3e^{-\frac{3}{5}x}$
$\frac{\mathrm{d} }{\mathrm{d} x}(e^{\frac{2}{5}x}y)=3e^{-\frac{3}{5}x}$
$e^{\frac{2}{5}x}y=-5e^{-\frac{3}{5}x}+c$
$y=e^{-x}(ce^{\frac{3}{5}x}-5)$
Let’s plug in the initial conditions to find the constant $c$: $5=e^{-0}(ce^{\frac{3}{5}\cdot 0}-5)$
$c=10$
$y=5e^{-x}(2e^{\frac{3}{5}x}-1)$
2. First Order Non-Exact Differential Equations
Use $\mu(x)=e^{\int \frac{\frac{\partial f}{\partial y}-\frac{\partial g}{\partial x}}{g} \mathrm{d}x}$ for when you have the form $f \mathrm{d}x+g \mathrm{d}y=0$, and the equation is not exact.
Like before, find $\mu(x)$ and multiply everything by $\mu(x)$.
2.1. Example
$(4 \sin y + \frac{6}{x})\mathrm{d}x+x\cos y \mathrm{d}y=0$
As you can see, the functions $f$ and $g$ are both functions of both $x$ and $y$, so we must use partial differentials. Don’t be scared - this just means you need to take the partial derivative of the function.
Using the formula, $f=4 \sin y + \frac{6}{x}$ and $g= x\cos y$.
Take the partial derivative of $f$ with respect to $y$, and the partial derivative of $g$ with respect to $x$. When you are taking the partial derivative of one variable, all the other variables are considered constants. $\frac{\partial f}{\partial y}=4\cos y$ and $\frac{\partial g}{\partial x}=\cos y$.
Therefore, $\mu(x)=e^{\int \frac{4\cos y -\cos y}{x \cos y}\mathrm{d}x}=x^3$.
Multiply every term by $\mu (x)$: $(4x^3 \sin y + 6x^2)\mathrm{d}x+x^4 \cos y \mathrm{d}y=0$.
Now this is just an exact equation you can solve. Let $F$ be the general solution to the equation and that $F=\frac{\partial F}{\partial x}\mathrm{d}x+\frac{\partial F}{\partial y}\mathrm{d}y=0$.
$f=\frac{\partial F}{\partial x}=4x^3 \sin y +6x^2$, $g=\frac{\partial F}{\partial y}=x^4 \cos y$
Using $\frac{\partial F}{\partial x}=4x^3 \sin y +6x^2$, integrate: $F=x^4 \sin y+2x^3+f(y)$.
Then differentiate: $\frac{\partial F}{\partial y}=x^4 \cos y+f’(y)$.
Compare: Therefore $f’(y)=0$, so $f(y)=c$.
Substitute into $F$: $F=x^4 \sin y+2x^3+c$, and recall that $F=0$, so $ x^4 \sin y+2x^3=c$.