Question

What is the antiderivative of `1/(x^2 - 2x +2)`?

Answer 1

`arctan(x-1)+C`

Explanation:

We want to find:

`intdx/(x^2-2x+2)`

You will want to recognize that this is close to fitting the form of the `arctan` integral, which is, for future reference:

`int(du)/(u^2+a^2)=1/aarctan(u/a)+C`

In order to write the denominator of the integral `x^2-2x+2`, we must complete the square. We will want to use the perfect square `x^2-2x+1=(x-1)^2`.

`=intdx/(x^2-2x+1+2-1)`

This is equal to the original integral expression—all that's been done is `+1` and then `-1` to balance it out.

Now, group the `(x^2-2x+1)` and `(2-1)`:

`=intdx/((x^2-2x+1)+(2-1))`

`=intdx/((x-1)^2+1)`

We can now apply the `arctan` integral, if we let

`{(u=x-1" "=>" "du=dx),(a=1):}`

This gives us

`=int(du)/(u^2+a^2)`

Which equals

`=1/aarctan(u/a)+C=1/1arctan((x-1)/1)+C`

`=arctan(x-1)+C`

Answer 2

We see that:

`intdx/(x^2-2x+2)=intdx/((x-1)^2+1)`

Let `x-1=tantheta`. This implies that `(x-1)^2+1=tan^2theta+1=sec^2theta` and that `dx=sec^2thetad theta`. Then the integral can be expressed as:

`=int(sec^2thetad theta)/sec^2theta=intd theta=theta+C`

From `tantheta=x-1` we see that `theta=arctan(x-1)`:

`=arctan(x-1)+C`