Clearly

$\frac{x+1}{\left(x-1\right)\left(x+2\right)}-\frac{x}{\left(x-1\right)\left(x-3\right)}>0$

$\Rightarrow \frac{-4x-3}{\left(x+2\right)\left(x-1\right)\left(x-3\right)}>0$

Then draw a straight line and mark $x$ values that make the fractors zero. That is $-2 ,\frac{3}{4} ,1,3$.

Then check the validity for some $x$ value less than $-2$.

Say $x=-3$.

For $x=-3$ we have that inequality does not hold.

So LHS is negative for $x=-3$

Then clearly $x \in \:\left(-2,\frac{3}{4}\right)\cup \left(1,\:3\right)$

Clearly

$\frac{x+1}{\left(x-1\right)\left(x+2\right)}-\frac{x}{\left(x-1\right)\left(x-3\right)}>0$

$\Rightarrow \frac{-4x-3}{\left(x+2\right)\left(x-1\right)\left(x-3\right)}>0$

Then draw a straight line and mark $x$ values that make the fractors zero. That is $-2 ,-\frac{3}{4} ,1,3$.

Then check the validity for some $x$ value less than $-2$.

Say $x=-3$.

For $x=-3$ we have that inequality does not hold.

So LHS is negative for $x=-3$

Then clearly $x \in \:\left(-2,-\frac{3}{4}\right)\cup \left(1,\:3\right)$

Clearly

$\frac{x+1}{\left(x-1\right)\left(x+2\right)}-\frac{x}{\left(x-1\right)\left(x-3\right)}>0$

$\Rightarrow \frac{-4x-3}{\left(x+2\right)\left(x-1\right)\left(x-3\right)}>0$

Then draw a straight line and mark $x$ values that make the fraction zero. That is $-2 ,\frac{3}{4} ,1,3$.

Then check the validity for some $x$ value less than $-2$.

Say $x=-3$.

For $x=-3$ we have that inequality does not hold.

So LHS is negative for $x=-3$

Then clearly $x \in \:\left(-2,\frac{3}{4}\right)\cup \left(1,\:3\right)$

Clearly

$\frac{x+1}{\left(x-1\right)\left(x+2\right)}-\frac{x}{\left(x-1\right)\left(x-3\right)}>0$

$\Rightarrow \frac{-4x-3}{\left(x+2\right)\left(x-1\right)\left(x-3\right)}>0$

Then draw a straight line and mark $x$ values that make the fractors zero. That is $-2 ,\frac{3}{4} ,1,3$.

Then check the validity for some $x$ value less than $-2$.

Say $x=-3$.

For $x=-3$ we have that inequality does not hold.

So LHS is negative for $x=-3$

Then clearly $x \in \:\left(-2,\frac{3}{4}\right)\cup \left(1,\:3\right)$