Integration with fraction I'm having difficulty understanding how below integral is evaluated : 

$$\int_0^1\frac{(1-y)^2}{2}+(1-y)y dy = \frac{-1(1-y)^3}{6}+\frac{y^2}{2}-\frac{y^3}{3}\bigg|^1_0$$

What are steps involved in this evaluation ?

For $\frac{(1-y)^2}{2}$ this appears to be evaluated as  $\frac{(1-y)^{2+1}}{(3)2}$ but I'm how unsure - 1 is inserted ?
Antiderivative if $(1-y)^{2}$ is not $(1-y)^{3} /3$. If you differentiate the latter you get $-(1-y)^{2}$. Hence the antiderivative is $-(1-y)^{3} /3$.
Remember that where $f(x)$ is linear we have that:

$$\int[f(x)]^n dx=\frac{[f(x)]^{n+1}}{(n+1)f'(x)}+C$$