What is the integral of sqrt((t^2)+1)?

Answer 1

let t = tan x, dt = sec^2 (x) dx
then 1+tan^2 (x) = sec^2 (x)

Int [sqrt(t^2 + 1)] dt
= Int { sqrt[sec^2 (x)] [sec^2(x)]} dx
= Int [sec^3 (x)] dx

Use Integration by part
Int [sec^3 (x)] dx
= sec x Int[ sec^2 (x)] dx
- Int [ (tan x)(sec x tan x)] dx
= (sec x)(tan x)
- Int [ (tan x)(sec x tan x)] dx ...(1)

From
(tan x)(sec x tan x)
= (sec x)[sec^2 (x) - 1]
= sec^3 (x) - sec x ...(2)

Replace (2) in (1)
Int [sec^3 (x)] dx
= (sec x)(tan x) - Int [ sec^3 (x) - sec x ] dx

2 Int [sec^3 (x)] dx
= (sec x)(tan x) - Int [ sec x ] dx

Int [sec^3 (x)] dx
= (1/2) [ (sec x)(tan x) - In(sec x + tan x)] ...(3)

Then replace sec x = Sqrt(1+t^2) and tan x = t in (3)

Answer 2

Use t = tan(θ), so dt = sec²(θ)dθ. You'll end up with the integral of sec³(θ)dθ, a classic whose integraion (involving integration by parts, with a twist) is in almost every calc book. You should get
½[ t*sqrt(1 + t²) + ln(t + sqrt(1+ t²)) ]

Answer 3

Figure it out yourself; try a trig substitution.