Integrate $\int \left(A x^2+B x+c\right) \, dx$ I am asked to find the solution to the initial value problem:
$$y'=\text{Ax}^2+\text{Bx}+c,$$
where $y(1)=1$,
I get:
$$\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d$$
But the answer to this is: 
$$y=\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1.$$
Could someone show me what has been done and explain why?

$$y'=\text{Ax}^2+\text{Bx}+c$$
$$y(x)=\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d \ \ \  (*) $$
Using the initial condition $y(1)=1$ we get:
$$y(1)=1 \Rightarrow \frac{A }{3}+\frac{B }{2}+c +d=1 \Rightarrow d=1-\frac{A}{3}-\frac{B}{2}-c$$
Replacing this at the relation $(*)$ we get:
$$y(x)=\frac{A x^3}{3}+\frac{B x^2}{2}+c x+1-\frac{A}{3}-\frac{B}{2}-c=\frac{A x^3-A}{3}+\frac{B x^2-B}{2} +cx-c+1 \\ \Rightarrow y(x)=\frac{1}{3}A(x^3-1)+\frac{1}{2}B(x^2-1) +c(x-1)+1 $$
