friends.
While solving the differential equation $ (2xy' + y) \cdot \sqrt{1+x} = 1+2x $, I was faced with the following integral: $ \int \frac{(1+2x) \cdot \sqrt{x}}{2x \cdot \sqrt{1+x}} dx $ and I have absolutely no idea how to solve it!
I tried using Symbolab to see the "how-to" but their answer is not very correct, so I turned to Wolfram|Alpha and got: $ \sqrt{x} \cdot \sqrt{x+1} + C $. However, Wolfram|Alpha does not have a "step-by-step" tool (I'm a free-user);
I've tried several resolution tools: variable changes, splitting the integral, etc.
Could anyone shed some light on my problem?
Much appreciated!
Kind regards, Pedro.
It looks worse than it is: cancel out a factor of $\sqrt{x}$ between the numerator and denominator to get
$$\int \frac{1+2x}{\sqrt{x} \sqrt{x+1}} dx = \int \frac{1+2x}{\sqrt{x^2+x}} dx.$$
Then just recognize $(x^2+x)'=1+2x$.