Question

How do we integrate `int 1/(x^2 - 2x +5) dx`?

Answer 1

The integral equals `1/2arctan((x- 1)/2) + C`

Explanation:

We first notice that an integration by partial fractions is not easily possible, because we cannot factor `x^2 - 2x + 5`.

Also, since the degree of the numerator differs from the degree of the denominator by more than `1`, a substitution cannot be considered either. So instead we complete the square in the denominator to see what we can do.

`I = int 1/(x^2 - 2x + 5)dx`

`I = int 1/(1(x^2 - 2x +1 - 1) + 5)dx`

`I = int 1/((x -1)^2 - 1 + 5)dx`

`I = int 1/((x - 1)^2 + 4)dx`

Now if we let `u = (x - 1)/2`, then `du = 2dx` and we can substitute as follows:

`I = 2int 1/((2u)^2 + 4)du`

`I =2 int 1/(4u^2 + 4) du`

`I = 2int 1/(4(u^2 + 1)) du`

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

This is now a known integral.

`I = 1/2arctanu + C`

`I = 1/2arctan((x- 1)/2) + C`

This is the final answer.

Hopefully this helps!