Integral Exercise 23

We are asked to compute the following integral:

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

This integral can be approached in more than one way.

Method 1

We begin by rewriting the integrand in a more convenient form:

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

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

Next, we add and subtract 1 in the numerator:

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

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

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

$$ = - \left[ \int 1 \ dx - 2 \int \frac{1}{x^2 + 1} \ dx \right] $$

$$ = -x + 2 \cdot \arctan(x) + c $$

So the solution is:

$$ \int \frac{1 - x^2}{x^2 + 1} \ dx = -x + 2 \cdot \arctan(x) + c $$

Alternatively, we can split the integrand directly:

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

Using the linearity of the integral, we separate the terms:

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

The first integral is straightforward:

∫ 1/(x² + 1) dx = arctan(x) + c

So we have:

$$ \arctan(x) + C - \int \frac{x^2}{x^2 + 1} \ dx $$

To evaluate the remaining integral, we rewrite the numerator as $ x^2 + 1 - 1 $:

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

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

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

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

$$ = \arctan(x) + C - x + \arctan(x) $$

$$ = 2 \cdot \arctan(x) - x + C $$

Which confirms the result obtained using the first method.

And so on.