what is the intergral from (0,2) for the function (x)/(x^2+5)^1/2?

Question

The anitderivative of the equation form the interval (0,2)

Answer 1

I = ∫ x / (x² + 5)^(1/2).dx
let u = x² + 5
du = 2x.dx
du/2 = x.dx
I = (1/2).∫ 1 / u^(1/2) du
I = (1/2). ∫ u^(-1/2).dx
I = u^(1/2)
I = (x² + 5)^(1/2) between limits 0 to 2
I = 3 - 5^(1/2)
I = 0.76

Answer 2

take t=x^2 + 5

dt/dx = 2x

integral for the function (x)/(x^2+5)^1/2 =
= integral x/t^1/2 dx = integral 1/2t^(1/2) dt =
= integral t^(-1/2)/2 + C = t^(1/2) + C

If x=2 then t=x^2+5=9 and t^(1/2) = 3
if x=0 then t=x^2+5=5 and t^(1/2) = sqrt(5)

So, the intergral from (0,2) for the function (x)/(x^2+5)^1/2 is 3-sqrt(5)

Answer 3

The easiest method will be , put :- t^2 = (x^2 +5)
tdt =xdx ; and integral wiil become:-
I =∫ t.dt/(t^2)^(1/2)
=∫ dt =t
= (x² + 5)^(1/2) between limits 0 to 2
I = 3 - 5^(1/2)
I = 0.764