Integral Exercise 18

We are asked to evaluate the following integral:

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

This integral can be approached using a few different techniques.

Method 1

We begin by factoring out the constant 2 from the denominator:

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

Which simplifies to:

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

Recognizing a squared term inside the denominator, we rewrite it as:

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

Let u = x / √2, so that:

$$ du = \frac{1}{\sqrt{2}} \ dx \quad \Rightarrow \quad dx = \sqrt{2} \ du $$

Substituting into the integral gives:

$$ \frac{1}{2} \int \frac{1}{1 + u^2} \cdot \sqrt{2} \ du $$

We factor out the constant $ \sqrt{2} $:

$$ \frac{\sqrt{2}}{2} \int \frac{1}{1 + u^2} \ du $$

Now simplify the coefficient:

$$ \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} $$

So the integral becomes:

$$ \frac{1}{\sqrt{2}} \int \frac{1}{1 + u^2} \ du $$

This is a standard form, as:

∫ 1 / (1 + u²) du = arctan(u) + C

Therefore:

$$ \frac{1}{\sqrt{2}} \arctan(u) + C $$

Substituting back $ u = \frac{x}{\sqrt{2}} $, we obtain:

$$ \frac{1}{\sqrt{2}} \arctan\left(\frac{x}{\sqrt{2}}\right) + C $$

This is the final result.

A more direct approach is to recognize the integral as a standard form and apply a known formula immediately.

We write:

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

Observe that the denominator can be rewritten as:

$$ 2 + x^2 = (\sqrt{2})^2 + x^2 = a^2 + x^2 $$

We now apply the standard identity for integrals of the form:

$$ \int \frac{1}{a^2 + x^2} \ dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C $$

where $ C $ is the constant of integration.

In this case, $ a = \sqrt{2} $, so we get:

$$ \int \frac{1}{2 + x^2} \ dx = \frac{1}{\sqrt{2}} \arctan\left(\frac{x}{\sqrt{2}}\right) + C $$

This gives us the same result more directly and efficiently.

And so on.