91Ó°ÊÓ

A rational function is a fraction where the numerator and the denominator are both polynomials. In simpler terms, it looks like an algebraic fraction. When dealing with rational functions, it's crucial to examine the degree of the polynomials in the numerator and denominator. For example, in the integral \( \int \frac{1}{t^{2}-3t+3} \, dt \), the denominator \( t^2 - 3t + 3 \) is a polynomial of degree two.
To integrate rational functions, one common approach is to use algebraic techniques such as long division or partial fraction decomposition. However, when the denominator is quadratic and doesn't factor easily, we might need to use completing the square, as well as trigonometric substitution or other methods.

Completing the square is a technique used to rewrite a quadratic polynomial in the form \( ax^2 + bx + c \) into a perfect square trinomial. This makes it easier to evaluate integrals involving quadratic expressions. In our case, the quadratic \( t^2 - 3t + 3 \) becomes easier to work with when rewritten as \( (t - \frac{3}{2})^2 + \frac{3}{4} \).
Here's how you complete the square:

• Take the coefficient of the linear term (here \(-3\)), halve it, and square it. You get \( \left(\frac{-3}{2}\right)^2 = \frac{9}{4} \).
• Add and subtract this square inside the polynomial: \( t^2 - 3t + \frac{9}{4} - \frac{9}{4} + 3 \).
• Simplify to get \( (t - \frac{3}{2})^2 + \frac{3}{4} \).

Completing the square transforms the quadratic into a form that is ideal for trigonometric substitution or applying the arctan formula.

Trigonometric substitution is a method used to simplify the integration of expressions involving square roots or other difficult terms, often after completing the square. It involves substituting trigonometric functions in place of variables to utilize trigonometric identities and simplify the integral.
In the context of our integral \( \int \frac{1}{u^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, du \), substituting may not be necessary since it already mirrors the form suitable for applying the arctan formula directly. In other contexts, a substitution might involve setting \( u = a \cdot \tan(\theta) \) which capitalizes on the identity: \( 1 + \tan^2(\theta) = \sec^2(\theta) \). This identity helps in rewriting and solving the integrals in a more manageable form.

The arctan formula is invaluable when solving integrals of the form \( \int \frac{1}{a^2 + x^2} \, dx \). This formula expresses the result as \( \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \), where \( C \) is the constant of integration.
In our exercise, the expression \( u^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \) perfectly aligns with the form needed for the arctan formula. Therefore, integrating \( \int \frac{1}{u^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, du \) becomes straightforward. We recognize \( a = \frac{\sqrt{3}}{2} \), plug it into the formula, and after integration, we find:
\[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2u}{\sqrt{3}}\right) + C \]
Finally, substitute back \( u = t - \frac{3}{2} \) to get the integral in terms of the original variable \( t \). This technique is a powerful tool for handling integrals involving quadratic expressions.