find value integral $\int_{1}^{2}f(x)dx$

Question

For a function $f(x)$ determined and continuous with $\forall x\in $ $\Bbb R\setminus\left\{ 0 \right\}$ such that

$$x^{2}f^{2}(x)+(2x-1)f(x)=xf'(x)-1$$ and $f(1)=-2$

Find $$\int_{1}^{2} f(x)dx $$

I write: $$(xf(x)+1)^{2}=(xf(x)+1)'$$

But I do not know how to proceed

Answer 1

Following from the comments, you can rewrite $g(x) = xf(x)+1$. Then

$$ g(1) = 1\cdot f(1) + 1 = -1 $$

The equation becomes

$$ \frac{g'(x)}{(g(x))^2} = 1 $$

Then $$ \left[-\frac{1}{g(x)}\right]' = 1 $$

Integrating both sides

$$ -\frac{1}{g(x)} = x + C $$

Using the initial condition as above, we have $$ -\frac{1}{g(1)} = 1 + C \implies 1 + C = 1 \implies C = 0 $$

Solving for $g(x)$

$$ g(x) = -\frac{1}{x} $$

Solving for $f(x)$

$$ f(x) = \frac{g(x)-1}{x} = -\frac{1}{x^2} - \frac{1}{x} $$

You can now do the integration

Answer 2

The easiest course of action seems to be to solve the ODE $(xf(x)+1)^{2}=(xf(x)+1)'$ rather than trying to use this property to compute the integral in some clever way. If we write $g(x)=xf(x)+1$, the ODE becomes $g(x)^2=g'(x)$ with $g(1)=-1$.

As noted in the comments, this is a separable ODE. The general solution is $g(x)=(C-x)^{-1}$ (where defined), and with $g(1)=-1$ we get $g(x)=-1/x$. I am leaving the details out; it is a good exercise to solve this, and you can always ask a follow-up question for help if you get stuck.

We thus found that $xf(x)+1=-1/x$, so that $f(x)=-x^{-1}-x^{-2}$. The integral is now straightforward to calculate.