$u$-substitution always evaluates to $0$ Consider the integral 
$$ \int_a^b f(x) \,dx $$
now make the $u$-substitution $u \mapsto c + (x-a)(x-b)$. The resulting integral is
$$ \int_c^c h(u) \,du $$
where $h(u)$ is the integrand $f$ after the substitution, however, regardless of $f$ the integral $\int_c^c du = 0$. Looking at the definition Wikipedia provides I believe the substitution meets every condition. It's differentiable and has a integrable derivative, because it's a polynomial. So what's wrong with this substitution such that it always results in $0$?

Assume $a<b$.
If $u=c+(x-a)(x-b)$. Then
\begin{align}
u&=x^2-(a+b)x+ab+c\\
&=\left(x-\frac{a+b}{2}\right)^2+ab-\left(\frac{a+b}{2}\right)^2+c\\
&=\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2+c\\
x&=\frac{a+b}{2}\pm\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}
\end{align}
When $\displaystyle x<\frac{a+b}{2}$, $\displaystyle x=\frac{a+b}{2}-\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}$ and
$$\frac{du}{dx}=2x-(a+b)=-\sqrt{(a-b)^2-4c-4u}$$
When $\displaystyle x\ge\frac{a+b}{2}$, $\displaystyle x=\frac{a+b}{2}+\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}$ and
$$\frac{du}{dx}=2x-(a+b)=\sqrt{(a-b)^2-4c-4u}$$
When $\displaystyle x=\frac{a+b}{2}$, $\displaystyle u=c-\left(\frac{a-b}{2}\right)^2$
Therefore,
\begin{align}
\int_a^bf(x)dx&=\int_a^{\frac{a+b}{2}}f(x)dx+\int_{\frac{a+b}{2}}^bf(x)dx\\
&=\int_0^{c-\left(\frac{a-b}{2}\right)^2}\left[\frac{f\left(\frac{a+b}{2}-\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}\right)}{-\sqrt{(a-b)^2-4c-4u}}\right]du\\
&\qquad +\int_{c-\left(\frac{a-b}{2}\right)^2}^0\left[\frac{f\left(\frac{a+b}{2}+\sqrt{\left(\frac{a-b}{2}\right)^2-c-u}\right)}{\sqrt{(a-b)^2-4c-4u}}\right]du
\end{align}
It looks quite complicated. But my point is that the definite integral should be broken into two parts with different integrands.
