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.

  • $\begingroup$ 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

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 $(x^2+x)'=1+2x$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.