Question

Evaluate:

`int_0^2 1/sqrt(x^2 + 2x + 3) dx`

Answer 1

Quadratic expression inside the square root: x² + 2x + 3

We can rewrite this as:

x² + 2x + 1 + 2 = (x + 1)² + 2

The integral becomes:

`int_0^2 1/((x + 1)^2 + 2) dx`

Let,

u = x + 1

Then,

du = dx

Now we need to update our integration limits:

When x = 0, u = 0 + 1 = 1

When x = 2, u = 2 + 1 = 3

Substituting these into the integral:

`int_1^3 1/sqrt(u^2 + (sqrt2)^2) dx`

Use the standard formula:

`int 1/sqrt(u^2 + a^2) du = In |u + sqrt(u^2 + a^2)| + C`

a = `sqrt2`

`[In |u + sqrt(u^2 + 2)|]_1^3`

Evaluate the upper and lower bounds:

In `(3 + sqrt(3^2 + 2) - In (1 + sqrt(1^2 + 2))`

In `(3 + sqrt11) - In (1 + sqrt3)`

Using the properties of logarithms (In a − In b = In `a/b`)

In `(3 + sqrt11)/(1 + sqrt3)`