OpenStudy (anonymous):

Integration! $\int \frac{x + 1}{\sqrt{x^2 +1}} dx$

4 years ago
OpenStudy (anonymous):

I am tempted to stick a $$u$$ in but I end up $$u = x^2 + 1$$ $$du = 2x$$ so.. I am not sure what to do with that + 1

4 years ago
OpenStudy (anonymous):

Oh, I guess I can break it in half. $\int \frac{x}{\sqrt{x^2+1}} + \frac{1}{\sqrt{x^2+1}}$

4 years ago
OpenStudy (anonymous):

ya u can with the squared of both halves

4 years ago
OpenStudy (anonymous):

@Starr_Dynasty I am pretty sure this problem is beyond your current knowledge. Thank you for looking, at least!

4 years ago
OpenStudy (anonymous):

lol i know u welcome i just want to help u like you've been helping me i just need help on a couple more and i wouldnt really bother u as much

4 years ago
OpenStudy (anonymous):

So on the left I can do: $u = x^2 + 1$$du = 2x dx$ And then $$\int \frac{1}{2\sqrt{u}} du$$ $\sqrt{u} + C$ Well that's boring.

4 years ago
OpenStudy (anonymous):

According to a big table of integrals$\int \frac{1}{\sqrt{x^2 + 1}} dx = \ln |x + \sqrt{x^2 +1}| + C$

4 years ago
OpenStudy (anonymous):

So I have $\sqrt{x^2 + 1} + \ln |x + \sqrt{x^2+ 1}| + C$ According to Wolfram I can clean up that ln mess as $$\sinh^{-1}x$$

4 years ago