help with integration?

Question

How do you integrare

sqRt(4x^2+4) ?

I knowt that the answer is

xsqrt(4x^2+4) +2ln(2x+sqrt(4x^2+4)) +C

it says so in the back of my textbook, but I am completely lost about how to get there.

Answer 1

∫√(4x^2+4)
∫√4(x^2+1)
2∫√(x^2+1)
Using trig substitution let x =tan Ө
dx/dӨ = sec^2(Ө)
dx = sec^2 (Ө) dӨ

Which gives you this integral
2∫(tan^2(Ө)+1) sec^2 (Ө) dӨ
2∫(sec^2(Ө)) sec^2(Ө)
2∫sec^3(Ө) dӨ

Integrate by parts..
2∫sec^3(Ө) dӨ
2∫sec(Ө)^2 sec(Ө) dӨ
u = sec(Ө)
dv = sec^2 (Ө)
du = sec(Ө) tan(Ө) dx
v = tan(Ө)

∫sec(Ө)^3 = sec(Ө) tan(Ө) - ∫tan(Ө) sec(Ө) tan(Ө)
∫sec(Ө)^3 = sec(Ө) tan(Ө) - ∫tan^2 (Ө) sec(Ө)

Break up the integral using tan^2 = sec^2 - 1
∫sec(Ө)^3 = sec(Ө) tan(Ө) - ∫(sec^2 (Ө)-1) sec(Ө)
∫sec(Ө)^3 = sec(Ө) tan(Ө) - ∫(sec^3 (Ө)- sec(Ө))
∫sec(Ө)^3 = sec(Ө) tan(Ө) - ∫sec^3(Ө) + ∫sec(Ө)
Add the ∫sec^3(Ө) to both side
2 ∫sec(Ө)^3 = sec(Ө) tan(Ө) + ∫sec(Ө)
2 ∫sec(Ө)^3 = sec(Ө) tan(Ө) + ln| sec(Ө) + tan(Ө) |

which is what we wanted to find. Not to put it back in terms of x. since x = tan(Ө)
we have a triangle with opposite side x, adjacent 1 (by definition of tangent. This makes the hypotenuse sqrt(1+x^2). So our sec(Ө)'s are sqrt(1+x^2) tan(Ө) is x

x sqrt(1+x^2) + ln | sqrt(1+x^2) + x | + C

Can simplify it a little more using the fact that arcsinh(x)= ln(x+sqrt(x^2+1))

x sqrt(1+x^2) + arcsinh(x) + C

Answer 2

xsqrt(4x^2+4) +2ln(2x+sqrt(4x^2+4)) +C

Answer 3

sqrt[x^2 + 1] + sinh^-1 (x) + C

No idea how to get there though... I used my integral website to do it for me. I am good at integrals, just can't hit it this time.

http://integrals.wolfram.com/index.jsp

for future problems u can't find answers to.