integration of 1/t^3 square root of t^2-1 From square root of 2 to 2

We'll start evaluating the integral, using
substitution.

Let sqrt(t^2 - 1) =
u.

We'll differentiate both
sides:

2tdt/2sqrt(t^2 - 1) =
du

du =tdt/sqrt(t^2 - 1)

We'll
determine t:

u = sqrt(t^2 -
1)

u^2 = t^2 - 1 => t^2 = u^2 +
1

t^4 = (u^2 + 1)^2

We'll
re-write the integral:

`int` dt/t^3*sqrt(t^2-1) = `int`
tdt/t^4*sqrt(t^2 - 1)

`int` tdt/t^4*sqrt(t^2 - 1) = `int`
du/(u^2 + 1)^2

`int` du/(u^2 + 1)^2 = `int` (u^2 + 1 -
u^2)dt/(u^2 + 1)^2

`int` (u^2 + 1 - u^2)dt/(u^2 + 1)^2 =
`int` dt - `int` u^2dt//(u^2 + 1)^2

We'll integrate the
last integral by parts, using the formula:

`int` f*g' = f*g
- `int` f'*g

Let f = u => f' =
du

Let g' = u/(u^2 + 1)^2 => g = `int` udu/(u^2 +
1)^2

We'll determine g using
substitution:

u^2 + 1 = v => 2udu = dv => udu
= dv/2

g = `int` dv/2v^2 = -1/2(u^2 +
1)

We'll apply the formula of integrating by
parts:

`int` u^2dt//(u^2 + 1)^2 = -u/2(u^2 + 1) + `int`
du/2(u^2 + 1)

`int` u^2dt//(u^2 + 1)^2 = -u/2(u^2 + 1) +
(1/2)*arctan u + C

The result of integration of the
function is:

`int` dt/t^3*sqrt(t^2-1) = sqrt(t^2 - 1)
+  sqrt(t^2 - 1)/2t^2 - (1/2)*arctan [sqrt(t^2 - 1)] + C

To
determine the value of the definite integral, we'll use Leibniz Newton
formula:

`int`
dt/t^3*sqrt(t^2-1) = F(2) -
F(sqrt2)

F(sqrt2) = sqrt(2 - 1) +  sqrt(2 - 1)/4 -
(1/2)*arctan [sqrt(2 - 1)]

F(sqrt2) = 1 +  1/4 - (`pi`
/8)

F(2) = sqrt(2^2 - 1) +  sqrt(2^2 - 1)/8 - (1/2)*arctan
[sqrt(2^2 - 1)]

F(2) = sqrt 3+  sqrt 3/8 - (1/2)*arctan
[sqrt3]

`int`
dt/t^3*sqrt(t^2-1) = F(2) - F(sqrt2) =
sqrt 3+  sqrt 3/8 - (1/2)*arctan [sqrt3]- 5/4 + `pi` /
8