Integral Calculation Exercise 17

In this exercise, we're asked to compute the integral

$$ \int \frac{1}{x^2 + x^4} \ dx $$

To begin, we factor the denominator:

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

The integrand is now expressed in terms of irreducible factors.

To proceed, we apply the method of partial fraction decomposition.

$$ \int \frac{1}{x^2 \cdot (1 + x^2)} \ dx = \int \left( \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{1 + x^2} \right) dx $$

To determine the constants, we rewrite the right-hand side as a single rational expression:

$$ \int \frac{A \cdot x \cdot (1 + x^2) + B \cdot (1 + x^2) + (Cx + D) \cdot x^2}{x^2 \cdot (1 + x^2)} \ dx $$

Expanding the numerator:

$$ \int \frac{Ax + Ax^3 + B + Bx^2 + Cx^3 + Dx^2}{x^2 \cdot (1 + x^2)} \ dx $$

Grouping like terms yields:

$$ \int \frac{x^3(A + C) + x^2(B + D) + xA + B}{x^2 \cdot (1 + x^2)} \ dx $$

We now equate this expression to the original integrand and compare numerators. Since the left-hand side has numerator 1, we obtain the system:

We can now solve this system of equations. Starting from A = 0, we find:

$$ \begin{cases} C = 0 \\ B + D = 0 \\ B = 1 \end{cases} $$

Substituting B = 1 into the second equation gives:

$$ D = -1 $$

Thus, the values of the constants are: A = 0, B = 1, C = 0, D = -1.

We now substitute these values back into the decomposition:

$$ \int \frac{1}{x^2 \cdot (1 + x^2)} \ dx = \int \left( \frac{0}{x} + \frac{1}{x^2} + \frac{0 \cdot x - 1}{1 + x^2} \right) dx $$

Which simplifies to:

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

Using the linearity of the integral, we write:

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

Both integrals on the right are elementary and can be computed directly.

The first is a power function:

$ \int x^{-2} \ dx = -x^{-1} + c $, which can be written as $ -\frac{1}{x} + c $.

The second is a standard arctangent integral:

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

Combining both results gives:

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

This is the final solution to the integral.

And so on.