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

Mar 22, 2016

$\arctan \left(x - 1\right) + C$

#### Explanation:

We want to find:

$\int \frac{\mathrm{dx}}{{x}^{2} - 2 x + 2}$

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

$\int \frac{\mathrm{du}}{{u}^{2} + {a}^{2}} = \frac{1}{a} \arctan \left(\frac{u}{a}\right) + C$

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

$= \int \frac{\mathrm{dx}}{{x}^{2} - 2 x + 1 + 2 - 1}$

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 $\left({x}^{2} - 2 x + 1\right)$ and $\left(2 - 1\right)$:

$= \int \frac{\mathrm{dx}}{\left({x}^{2} - 2 x + 1\right) + \left(2 - 1\right)}$

$= \int \frac{\mathrm{dx}}{{\left(x - 1\right)}^{2} + 1}$

We can now apply the $\arctan$ integral, if we let

$\left\{\begin{matrix}u = x - 1 \text{ "=>" } \mathrm{du} = \mathrm{dx} \\ a = 1\end{matrix}\right.$

This gives us

$= \int \frac{\mathrm{du}}{{u}^{2} + {a}^{2}}$

Which equals

$= \frac{1}{a} \arctan \left(\frac{u}{a}\right) + C = \frac{1}{1} \arctan \left(\frac{x - 1}{1}\right) + C$

$= \arctan \left(x - 1\right) + C$

Jun 22, 2017

We see that:

$\int \frac{\mathrm{dx}}{{x}^{2} - 2 x + 2} = \int \frac{\mathrm{dx}}{{\left(x - 1\right)}^{2} + 1}$

Let $x - 1 = \tan \theta$. This implies that ${\left(x - 1\right)}^{2} + 1 = {\tan}^{2} \theta + 1 = {\sec}^{2} \theta$ and that $\mathrm{dx} = {\sec}^{2} \theta d \theta$. Then the integral can be expressed as:

$= \int \frac{{\sec}^{2} \theta d \theta}{\sec} ^ 2 \theta = \int d \theta = \theta + C$

From $\tan \theta = x - 1$ we see that $\theta = \arctan \left(x - 1\right)$:

$= \arctan \left(x - 1\right) + C$