Integrate a quotient with fractional power of a quadratic polynomial I need help finding the indefinite integral of
$$\int\,\frac{x}{(7x - 10 - {x^2})^{3/2}}\,\text{d}x\,.$$
 A: HINT:
As $$7x-10-x^2=-\dfrac{4x^2+40-28x}4=\dfrac{3^2-(2x-7)^2}4$$
Set $2x-7=3\sin\theta$
See: Trigonometric substitutions 
A: $$\int { \frac { x }{ \sqrt { { \left( 7x-10-{ x }^{ 2 } \right)  }^{ 3 } }  } dx } =\int { \frac { x }{ \sqrt { { -\left( { x }^{ 2 }-7x+10 \right)  }^{ 3 } }  } dx= } \int { \frac { x }{ \sqrt { { -\left( { x }^{ 2 }-7x+\frac { 49 }{ 4 } -\frac { 49 }{ 4 } +10 \right)  }^{ 3 } }  } dx= } \\ =\int { \frac { x }{ \sqrt { { { \left( \frac { 9 }{ 4 } -{ \left( x-\frac { 7 }{ 2 }  \right)  }^{ 2 } \right)  }^{ 3 } } }  } dx } \\ x-\frac { 7 }{ 2 } =\frac { 3 }{ 2 } \sin { t } \\ x=\frac { 1 }{ 2 } \left( 7+3\sin { t }  \right) \Rightarrow dx=\frac { 3 }{ 2 } \cos { t } dt\\ \int { \frac { \frac { 3 }{ 2 } \cos { t }  }{ \sqrt { { { \left( \frac { 9 }{ 4 } -{ \frac { 9 }{ 4 }  }\sin ^{ 2 }{ t }  \right)  }^{ 3 } } }  } dt } =\int { \frac { \frac { 3 }{ 2 } \cos { t }  }{ { \left( \frac { 3 }{ 2 }  \right)  }^{ 3 }\sqrt { { { \left( 1-\sin ^{ 2 }{ t }  \right)  }^{ 3 } } }  } dt } =\frac { 4 }{ 9 } \int { \frac { \cos { t }  }{ \cos ^{ 3 }{ t }  } dt=\frac { 4 }{ 9 } \int { \frac { dt }{ \cos ^{ 2 }{ t }  } =\frac { 4 }{ 9 }  }  } \tan { t } +C\\ \sin { t } =\left( \frac { 2x-7 }{ 3 }  \right) \Rightarrow t=\arcsin { \left( \frac { 2x-7 }{ 3 }  \right)  } \\ \frac { \\ 4 }{ 9 } \tan { \arcsin { \left( \frac { 2x-7 }{ 3 }  \right)  }  } +C=\frac { 4 }{ 9 } \frac { \left( \frac { 2x-7 }{ 3 }  \right)  }{ \sqrt { 1-{ \left( \frac { 2x-7 }{ 3 }  \right)  }^{ 2 } }  } +C=\frac { 8x-28 }{ 9\sqrt { -4x^{ 2 }+28x-40 }  } +C$$
so

$$\int { \frac { x }{ \sqrt { { \left( 7x-10-{ x }^{ 2 } \right)  }^{ 3 } }  } dx } =\frac { 8x-28 }{ 9\sqrt { -4x^{ 2 }+28x-40 }  } +C$$

A: 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.
