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

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### 2 Answers

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)]**

To evaluate Integral f(x) dx x = 0 to1.

f(x) = 1/(x^2+2x+5)

Solution:

f(x) = 1/(x+1)^2 +4

Put x+1 = t. Then dx = dt, when x= 0 to 1, t= 1 to 2.

{Integral f(x) dx , x=0 to 1} = {integral dt/(t+2^2) , t = 1 to 2.

= (1/2){arc tan t/2 , t= 1, 2}, as Integral dx/(x^2+a^2) = (1/a) arctan (x/a).

= {(1/2) arc tan (2/2) - (1/2) arc tan (1/2)}

= (1/2{ arctan1 - arctan(1/2)}

= (1/2){pi/4 - aerctan(1/2)}

= (1/2){0.785398163 - 0.463647609}

= 0.160875277 radians.