If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f (x)$ 
If  $f (x) +f'(x) = x^3+5x^2+x+2$  then  find $f(x)$.

$f'(x)$ is the first derivative of $f (x)$.
I have no idea about this question, please help me. 

The hard way:
You can use the fact that
$$(e^xf(x))'=e^x(f(x)+f'(x)).$$
Then
$$(e^xf(x))'=e^x(x^3+5x^2+x+2).$$
Now, by integration (which can be performed by parts, integrating $e^x$),
$$e^xf(x)=\int e^x(x^3+5x^2+x+2)dx=e^x(x^3+2x^2-3x+5)+C$$ and
$$f(x)=x^3+2x^2-3x+5+Ce^{-x}.$$
[I doubt this is the method you are expected to use...]

Empirically:
Let us assume that the solution is close to $f_0(x)=x^3$ and let us plug in the given equation.
$$f_0(x)+f'_0(x)=x^3+3x^2\ne x^3+5x^2+x+2.$$ But we are off by the polynomial $$2x^2+x+2$$ which is of a lower degree.
So let us try $f_1(x)=x^3+2x^2$, giving
$$f_1(x)+f'_1(x)=x^3+5x^3+4x\ne x^3+5x^2+x+2.$$
This time, we are only off by $-3x+2$, one degree less.
$$f_2(x)=x^3+2x^2-3x$$ then $$f_2(x)+f'_2(x)=x^3+5x^2+x-3.$$
And finally
$$f_3(x)=x^3+2x^2-3x+5$$ has converged to a solution.

Epilogue:
Let us assume that there exists another solution $f$ which is different from $f_3$.
From
$$f(x)+f'(x)=f_3(x)+f'_3(x)= x^3+5x^2+x+2,$$ we can draw that
$$f(x)-f_3(x)+f'(x)-f'_3(x)=0$$ which is of the form
$$g(x)+g'(x)=0.$$
It is not difficult to show that no polynomial is the opposite of its derivative. Actually, there is a single differentiable function having this property: the negative exponential
$$g(x)=e^{-x}$$ (or multiples).
Grouping these results,
$$f(x)=x^3+5x^2+x+2+Ce^{-x}.$$
