Question

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

Answer 1

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

Answer 2

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

Answer 3

ya u can with the squared of both halves

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

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 \(\sinh^{-1}x\)