Show that $Y^2 + X^2(X+1)^2$ is irreducible over $\mathbf R$ 
Show that $Y^2 + X^2(X+1)^2$ is irreducible over $\mathbf R$.

Are there some general tricks for avoiding barbaric computations in general case?

If it were reducible it would factor in two non-unit factors. Either both of them are of degree 1 in $Y$ or one of them is of degree zero in $Y$ and the other is of degree 2 in $Y$. 
In the second case we see that the factor that is of degree zero in $Y$ must divide $Y^2$. Therefore it must be a unit, which is a contradiction.
Assume then that the two factors are of degree one in $Y$. So
$$Y^2+X^2(X+1)^2=(A(X)Y+B(X))(C(X)Y+D(X))=A(X)C(X)Y^2+(A(X)D(X)+C(X)B(X))Y+B(X)D(X)$$
From this $A(X)$ and $C(X)$ are units. So, we can assume they are $1$. We get
$$Y^2+X^2(X+1)^2=(Y+B(X))(Y+D(X))=Y^2+(B(X)+D(X))Y+B(X)D(X)$$
From this $B(X)+D(X)=0$ and $X^2(X+1)^2=B(X)D(X)$. Therefore 
$$X^2(X+1)^2=-B^2(X).$$
Therefore, 
$$1=-B^2_{0}$$
where $B_0$ is the leading term of $B(X)$. But there is no such $B_0$ in the reals, who's square is $-1$.
