Approximate integral: $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx.$ 
Approximate integral: $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx.$

My attempt:
Let I = $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx$
$u=7-x\implies I=\frac 72 \int_3^4 \frac 1{\sqrt{(7x-10-x^2)^3}}dx$ Now I have to approximate this... 
I got to this point too:
$u=3-x \implies I= \int_0^1 \frac {1}{\sqrt{(-x^2-x+2)^3}}dx.$
and: $\frac 1{-x^2-x+2}\leq1\implies \int_0^1 \frac{1}{\sqrt{(-x^2-x+2)^3}}dx\leq \int_0^1-\frac{1}{(x+2)(x-1)}dx$ and the right one does not converge...
I have these  values:
$a) (\frac 9{10};\frac {19}{20})$
$b) (\frac {19}{20},1)$
$c) (1;\frac {21}{20})$
$d) (\frac {21}{20};\frac {11}{10})$
$e) (\frac {11}{10},\frac {23}{20})$
$f) (\frac {23}{20}; \frac 65)$.

I will try to solve the problem.
The integrand has the form $P_1 \cdot Q_2 ^ {- 3/2} $.
It is logical to assume that the antiderivative has the form $ R_1 \cdot Q_2 ^ {- 1/2} $.
Differentiate and equate.
$\bigg((ax+b)(7x-10-x^2)^{-\frac{1}{2}}\bigg)'=x\cdot (7x-10-x^2)^{-\frac{3}{2}}$
So $\displaystyle a(7x-10-x^2)^{-\frac{1}{2}}+(ax+b)(7-2x)\cdot -\frac{1}{2}\cdot (7x-10-x^2)^{-\frac{3}{2}}=x\cdot (7x-10-x^2)^{-\frac{3}{2}}$
So $\displaystyle a(7x-10-x^2)+(ax+b)(7-2x)\cdot -\frac{1}{2}=x$
So we get $\displaystyle a=\frac{14}{9}$ and $\displaystyle b=-\frac{40}{9}$
So $\displaystyle \int^{4}_{3}\frac{x}{\sqrt{(7x-10x-x^2)^3}}dx=-\frac{1}{9}\bigg(\frac{14x-40}{\sqrt{7x-10-x^2}}\bigg)\bigg|^{4}_{3}=\frac{7\sqrt{2}}{9}.$
