Integrating $\sin(x)x^2$ by parts, why do we only add $C$ at the end?
The familiar formula for integration by parts is $$\int udv= uv-\int vdu $$
Now if you like to add a constant to your $v$ you get $$\int udv= u(v+c)-\int (v+c)du = uv+uc-\int vdu -c\int du = $$
$$uv+uc-\int vdu -cu = uv- \int vdu $$
Which is exactly the same result due to cancelation of $cu$ and -$cu$
Let's keep those $C_{i}$'s in those steps and show you that things still cancel out:
$$\int{x^{2}sinxdx}=-x^{2}cosx+C_{1}x^{2}-2C_{1}\int{xdx}+2\int{xcosxdx}$$
We get $$\int{x^{2}sinxdx}=-x^{2}cosx+C_{1}x^{2}-2C_{1}\int{xdx}+2xsinx+2C_{2}x-2\int{sinxdx}-2C_{2}\int{dx}$$
Then $$\int{x^{2}sinxdx}=-x^{2}cosx+C_{1}x^{2}-C_{1}(2)\frac{x^{2}}{2}+C_{3}+2xsinx+2xC_{2}+2cosx+C_{4}-2C_{2}x+C_{5}$$
Collection of like terms and letting $C=C_{3}+C_{4}+C_{5}$, we get $$\int{x^{2}sinxdx}=-x^{2}cosx+2xsinx+2cosx+C$$
For $f $ and $ g$ $ C^1$ at some intervalle,
The integration by parts is based on the identity $$(fg)'=f'g + fg'$$
which yields to
$$\int f'g = fg - \int fg'$$
If you write
$$\int f'g = (f+C_1)g - \int fg'$$ the result will be false.
You should write
$$\int f'g = (f+C_1)g - \int (f+C_1)g'$$ To satisfy the first identity.