integrate sqrt(1+x^2)?

Question

please show your work..

Answer 1

First, we integrate by parts:
u=√(1+x^2), v=x, du=x/√(1+x²) dx, dv=dx
∫√(1+x²) dx = x √(1+x²) - ∫x²/√(1+x²) dx
Rewriting the integral on the right:
∫√(1+x²) dx = x √(1+x²) - (∫(x²+1)/√(1+x²) dx - ∫1/√(x²+1) dx)
Simplifying:
∫√(1+x²) dx = x √(1+x²) - ∫√(x²+1) dx + ∫1/√(x²+1) dx
2∫√(1+x²) dx = x √(1+x²) + ∫1/√(x²+1) dx
2∫√(1+x²) dx = x √(1+x²) + ∫1/√(1-(ix)²) dx
Let sin u=ix, cos u du=i dx
2∫√(1+x²) dx = x √(1+x²) - ∫i/√(1-sin² u) cos u du
2∫√(1+x²) dx = x √(1+x²) - ∫i cos u/√(cos² u) du
2∫√(1+x²) dx = x √(1+x²) - iu + C
2∫√(1+x²) dx = x √(1+x²) - i arcsin (ix) + C
∫√(1+x²) dx = (x √(1+x²) - i arcsin (ix))/2 + C

At this point we are technically done, but let's see if we can remove those complex numbers. Let u=arcsin (ix). Then:

sin u = ix
-i sin u = x
-i sin (i(-iu))=x
sinh (-iu)=x (using the identity sinh x=-i sin (ix))
-iu = arcsinh x
So:
-i arcsin (ix)=arcsinh (x)
And:

∫√(1+x²) dx = (x √(1+x²) + arcsinh x)/2 + C

Answer 2

Make the following substitution.

Let u=tan(x). Use the fact that tan^2(x)+1=sec^2(x). And then from here on you should be able to do it.

Answer 3

... and finally, verify using Wolfram Integrator