what is the integral of rad (x^2+1)??? can someone show me the steps? quick!?

Question

umm, thanks but im seeking a simpler method using du...u know, like the u of f(x) is x squared plus one, so du = 2x dx, but then im stuck. i dont know how to substitute 2x dx into the equation

Answer 1

∫ √(x²+1) dx.
When you have the integral which is the
square root of a quadratic, you make a trig
substitution to get rid of the radical.
Remember that 1 + tan² t = sec² t,
so in this case we let x = tan t dx = sec² t dt
and our integral reduces to computing
∫ sec³ t dt.
Here we let
U = sec t dV = sec² t dt
dU = sec t tan t dt V = tan t
So ∫ sec³ t dt = sec t tan t - ∫ sec t tan² t dt
= sec t tan t - ∫ (sec² t -1)*sec t dt.
= sec t tan t + ln| sec t + tan t| - ∫ sec³ t dt.
Solving for ∫ sec³ t dt, we finally get
∫ sec³ t dt = ½[ sec t tan t + ln| sec t + tan t|].
Now we have to back substitute:
x = tan t, √(x²+1) = sec t and we get
∫ √(x²+1) dx = ½[ x √(x²+1) + ln| x + √(x²+1)| ]+ C.
I noticed you were looking for a "u" substitution.
u = x²+1. Good for you to try simpler things first!
Let's try it and see what happens:
Let u = x² + 1
Then x = √(u-1)
dx = 1/ 2√(u-1)
Then we get
½ ∫ √u du/ √(u-1),
which, IMHO, doesn't seem any easier than
what we started with.

Answer 2

trigonometric substitution, like x=tan dx=sec^2dθ, which means sec^2θ is in denominator, and ∫√tan^2+1/sec^2 becomes by pythagoras identity ∫cosθdθ

Answer 3

put it ur graphing calc
graph it
use the trace function and "trace" where the parabola crosses the x axis...it should be twice