irrational integral $ \int \frac{1+\sqrt{x^2+3x}}{2-\sqrt{x^2+3x}}\, dx$

Question

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?

Answer 1

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 $

Answer 2

In your integral ,you can make break the numnerator in terms of denominator by first putting a -ve sign in front of the integral , then proceed .

Answer 3

Just a note, whenever doing an integral of this form, recall that

$$\int \frac{P}{Q+\sqrt{R}}dx=\int \frac{P(Q-\sqrt{R})}{Q^2-R}dx.$$

Answer 4

Simplify the integrand first before enforcing the substitution to lighten the workload in expanding into partial fractions.

$$I = \int \frac{1+\sqrt{x^2+3x}}{2-\sqrt{x^2+3x}}\, dx \\ = -\int \frac{\left(2-\sqrt{x^2+3x}\right) - 3}{2-\sqrt{x^2+3x}}\, dx \\ = \int \left(\frac3{2-\sqrt{x^2+3x}} - 1\right)\,dx$$

Now with $t=\sqrt{x^2+3x}-x\iff x= \dfrac{t^2}{3-2t}$ and $dx=-2\dfrac{t^2-3t}{(3-2t)^2}\,dt$, the first term is

$$\int \frac3{2-\sqrt{x^2+3x}}\,dx = 6 \int \frac{t(t-3)}{(t-6)(t-1)(2t-3)} \,dt$$