Exact solution and Euler method approximation for a first order differential equation Having trouble wrapping my brain around this question.. don't know where to start.

A. Consider the differential equation  $f'(x) = (x + 1)f(x)$ with $f(0) =1$. What is the exact formula for $f(x)$?

I tried solving for $f(x)$ but that just gives me $(x+1)f'(x)$. I feel like that's not sufficient here.

B. Solve for $f(.2)$ using Eulers approximation method with increment $h = .01$ for $x\in[0,0.2]$.

Edit- my work for part A. 
integral f'(x) = integral(x+1) *f(x)+(x+1) * integral (f(x)
so f(x) = ((x^2+x)/2)*f(x) + (x+1) * (f(x)^2)/2. 
I don't believe this is the right answer. 

$$f'(x) = (x + 1)f(x)$$
Is a separable differential equation of first order.
Rewrite as:
$$ \frac {df}{dx}=(x+1)f$$
$$ \frac {df}{f}=(x+1)dx$$
Integrate both sides and apply initial condition.
