How to find prime factors of big numbers made up of big prime factors? How do I quickly find the prime factors of big numbers with difficult factors?
What I mean by difficult factors is that the usual factors like $2,3,5,7,11$ may not be present.
For eg: the prime factors of $4181$ are $37$ and $113$. 

How do I find them quickly?

 A: There are many ways to factor numbers.  One approach I like is the Lehmer sieve which is a mechanical device that tries to find a pair of numbers $x,y$ such that $n=x^2-y^2$, hence $n=(x-y)(x+y)$ gives a factorization.
A naive version is looking for an $x$ such that $x^2-n$ is a perfect square.  We get a lower bound $x\geq \sqrt{4181}>64$.  For $x=65,66,\ldots$, we see if $\sqrt{x^2-n}$ is an integer, and $x=75$ is the first (after some effort).  $\sqrt{75^2-4181}=38$, so we get factors $4181=37\cdot 113$, which we would need to verify are prime.
We can speed this calculation up by seeing when $x^2-n$ is a quadratic nonresidue modulo various primes.  For example, $x^2-4181$ is a square modulo $3$ exactly when $x$ is a multiple of $3$, and it is a square modulo $5$ exactly when $x\equiv 0,\pm 1\pmod{5}$.  Putting these together, $x$ must be $0,6,9\pmod{15}$.  Even with these few considerations, the $x$'s we would have tried above would only have been $x=66,69,75$ before finding a partial factorization.
A: Pollard's method works well for not too large numbers and it's a simple algorithm that doesn't require a lot of work to implement. You just need a calculator and do some arithmetic with it. This method is based on Fermat's little theorem, which states that:
$$a^{p-1} = 1\bmod p\tag{1}$$
where $a\neq 0 \bmod p$, and $p$ is prime number. We can then attempt to use Eq. (1) to find a prime factor $p$ of a composite number $N$, by computing powers of a number (commonly taken to be $2$) modulo $N$ and raise the output again to another power and iterate that process. The idea is then that computing modulo $N$ is consistent with computing modulo $p$ and at some point the outputs will get stuck at $1 \bmod p$. If that's the case then the output minus $1$ will be zero modulo $p$, which means that the output minus $1$ that we get will be some multiple of $p$. We can then take the GCD of the output minus $1$ and $N$ to find $p$.
Now, for the power to get stuck at $1$ means that the power needs to include some of the divisors of $p-1$ (note that while the $p-1$th power of $a$ equals $1$ modulo $p$, the least power that will yield $1$ can be any divisor of $p-1$). This means that we should start by raising $2$ to the power of powers of small primes that are likely present in the prime factorization of $p-1$.
In some cases you'll find that the power you've computed modulo $N$ itself becomes $1$, and that then allows you to extract $p$ via the exponent of $2$.
For $N = 4181$, let's start with:
$$2^{2^2}\bmod N = 16$$
This covers a factor of $4$ in $p-1$. Let's now deal with a factors of $3$, two of such factors should be good enough. The numbers have to remain exact integers when doing the computations, so we need to do this in steps. We have:
$$16^3\bmod N = -85$$
and
$$(-85)^3\bmod N = 482$$
Let's now cover a factor of $5$ in $p-1$:
$$482^5\bmod N =482^4\times 482$$
$$482^2\bmod N =2369$$
$$482^4\bmod N = 2369^2\bmod N = 1259$$
$$482^5\bmod N = 1259\times 482\bmod N= 593$$
Let's now do a test with this number. We then need to compute the GCD of $N$ with $592$ using Euclid's algorithm. This algorithm works by exploiting the fact that the GCD of two numbers doesn't change if we add a multiple of one number to the other number. This means that we can repeatedly reduce one number modulo the other number. This process will end when the smallest number becomes $0$, the GCD is given by the other number. Applying the algorithm yields:
$$\gcd(4181,592) = \gcd(592, 37) = \gcd(37,0) = 37$$
So, $37$ is a prime factor. We see that we didn't need to compute the fifth power as $5$ is not a factor of $36$, but since the output gets stuck at $1$ modulo $p$ there is no harm in going further than you need to go. 
