Evaluate the definite integral of f=1/(x^2+2x+5) from 0 to 1 .

To evaluate the definite integral, we'll re-write the
denominator.

x^2+2x+5 =
x^2+2x+4+1

We'll group the terms: x^2+2x+1 =
(x+1)^2

The denominator will
become:

(x+1)^2 + 4

We'll
write the integral:

Int f(x)dx = Int
dx/(x^2+2x+5)

Int dx/(x^2+2x+5) = Int dx/[(x+1)^2 +
4]

We'll substitute x+1 by
t.

x+1 = t

(x+1)' =
1*dx

t' = dt

So, dx =
dt

We'll re-write the integral in
t:

Int dx/[(x+1)^2 + 4] = Int
dt/(t^2+4)

We'll factorize by 4, at
denominator:

Int dt/4(t^2/4 + 1) = (1/4)*Int dt/[(t/2)^2 +
1]

(1/4)*Int dt/[(t/2)^2 + 1] = 2/4*arctg (t/2) =
(1/2)*arctg (t/2)

But t =
x+1

For x = 0 => t =
1

For x = 1 => t =
2

Now, we'll apply Leibniz-Newton
formula:

Intdx/(x^2+2x+5), from 0 to 1 = F(2) -
F(1)

F(2) = (1/2)*arctg (2/2) =
(1/2)*(pi/4)

F(1) = (1/2)*arctg
(1/2)

F(2) - F(1) = (1/2)*[pi/4 - arctg
(1/2)]

Intdx/(x^2+2x+5), from 0 to 1 =
(1/2)*[pi/4 - arctg (1/2)]