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:
$a) (\frac 9{10};\frac {19}{20})$
$b) (\frac {19}{20},1)$
$c) (1;\frac {21}{20})$
$d) (\frac {21}{20};\frac {11}{10})$
$e) (\frac {11}{10},\frac {23}{20})$
$f) (\frac {23}{20}; \frac 65)$.

$$ \int_{3}^{4}\frac{x}{\sqrt{(x-2)^3 (5-x)^3}}\stackrel{x\mapsto\frac{7}{2}+z}{=}\int_{-1/2}^{1/2}\frac{z+7/2}{\sqrt{\left(9/4-z^2\right)^3}}\,dz=\frac{7}{2}\int_{-1/2}^{1/2}\frac{dz}{\sqrt{(9/4-z^2)^3}} $$
since the integral of an odd integrable function over a symmetric interval (with respect to the origin) equals zero. By symmetry again, the RHS equals
$$ 7\int_{0}^{1/2}\frac{dz}{\sqrt{(9/4-z^2)^3}}=28\int_{0}^{1}\frac{du}{\sqrt{(9-u^2)^3}}=\frac{28}{9}\int_{0}^{1/3}\frac{dv}{(1-v^2)^{3/2}} $$
or 
$$ \frac{28}{9}\int_{0}^{\arcsin(1/3)}\frac{d\theta}{\cos^2\theta}=\frac{28}{9}\tan\arcsin\frac{1}{3}=\color{blue}{\frac{7}{9}\sqrt{2}}\approx 1.09994. $$
This also has a nice geometric/probabilistic interpretation, since $\frac{\mathbb{1}_{(0,1)}(x)}{2\sqrt{x}}$ is the PDF of a squared $U(0,1)$ random variable and the given integral essentially is a convolution integral.
