<s>[INST] irrational integral $ \int \frac{1+\sqrt{x^2+3x}}{2-\sqrt{x^2+3x}}\, dx$ I have to solve this irrational integral $$ \int \frac{1+\sqrt{x^2+3x}}{2-\sqrt{x^2+3x}}\, dx$$
It seems that the most convenient way to operate is doing the substitution
$$ x= \frac{t^2}{3-2t}$$
according to the rule,
obtaining the integral:
$$ \int \frac{2 t^4-16t^3+36t^2-18t}{(t-2)(t+3)(3-2t)^2}\, dx$$
Then $$\frac{2 t^4-16t^3+36t^2-18t}{(t-2)(t+3)(3-2t)^2}= \frac{A}{t-2}+\frac{B}{t+3}+\frac{C}{3-2t}+\frac{D}{(3-2t)^2}$$
then I found the coefficient $A= \frac{12}{5},B= -\frac{12}{5}, C= \frac{15}{2},D= -\frac{9}{2}$
In my book the integral on which to operate is:
$$ \int -2\frac{(t^2-3t)(3+t-t^2)}{(t^2-7t+6)(3-2t)^2}\, dx$$
that is different from my
$$ \int \frac{2 t^4-16t^3+36t^2-18t}{(t-2)(t+3)(3-2t)^2}\, dx$$
Perhaps I made mistakes in the first passages and I checked lots of times my calculations. Can someone indicate where I'm making mistakes?
 [/INST] Note that the substitution $x=\frac{t^2}{3-2t}$ leads to
$ \sqrt{x^2+3x}=\frac{3t-t^2}{3-2t}, \>\>\>\>\> dx = -\frac{2(t^2-3t)}{(3-2t)^2}dt
$
and
$\int \frac{1+\sqrt{x^2+3x}}{2-\sqrt{x^2+3x}}dx
= -\int \frac{1+\frac{3t-t^2}{3-2t}}{2-\frac{3t-t^2}{3-2t}}\, \frac{2(t^2-3t)}{(3-2t)^2}dt
=  -\int \frac{3+t-t^2}{t^2-7t+6}\, \frac{2(t^2-3t)}{(3-2t)^2}dt
$
</s>