Integrate a quotient with fractional power of a quadratic polynomial I need help finding the indefinite integral of
$$\int\,\frac{x}{(7x - 10 - {x^2})^{3/2}}\,\text{d}x\,.$$

We have
$$
\int \frac{x}{(7x - 10 - x^2)^{3/2}}\,dx.
$$
Let $u=-x^2 +7x-10$, so that $du = (-2x+7)\,dx$ and $\dfrac{-du} 2 = \left( x - \dfrac 7 2 \right)\, dx$. Then the integral becomes
$$
\int\frac{x - \frac 7 2}{(7x-10+x^2)^{3/2}} \,dx + \frac 7 2 \int \frac{dx}{(7x-10+x^2)^{3/2}}
$$
The substitution above handles the first of these two integrlas; the one below handles the second:
\begin{align}
& -x^2+7x-10 = \overbrace{- (x^2-7x)-10 = -\left( x^2-7x + \frac{49}4 \right)^2 -10 + \frac{49}4}^{\text{completing the square}} \\[10pt]
= {} & -\left( x - \frac 7 2 \right)^2 + \left(\frac 3 2 \right)^2 = \left( \frac 3 2\right)^2 \left( 1 - \left( \frac 2 3 \left( x - \frac 7 2 \right) \right)^2 \right) \\[10pt]
= {} & \left( \frac 3 2 \right)^2 \left( 1 - \left( \frac{2x-7} 3 \right)^2 \right) = \left( \frac 3 2 \right)^2 (1-\sin^2\theta)
\end{align}
So
\begin{align}
\frac{2x-7} 3 & = \sin\theta \\[10pt]
\frac 2 3 \, dx & = \cos\theta\,d\theta
\end{align}
etc.
