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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.

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  • $\begingroup$ The presence of both $\sqrt x$ and $\sqrt{x+1}$ make a hyperbolic or trigonometric substitution of the form $x=\sinh^2t$ or $x=\tan^2u$ very tempting. $\endgroup$ – Lucian Oct 23 '15 at 4:36
  • $\begingroup$ That did show up when I tried using Symbolab as a resource, however, not having studied hyperbolic functions I decided to put it aside. How it would go using the $ x = sinh ^2 t $ substitution? I gotta confess I can't see how one can use the substitution $ x = \tan^2 u $ . Thanks for the interest, @Lucian $\endgroup$ – Pedro Cunha Oct 23 '15 at 20:48
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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 $(x^2+x)'=1+2x$.

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  • $\begingroup$ Brilliant. Thank you, sir. $\endgroup$ – Pedro Cunha Oct 23 '15 at 1:44

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