91Ó°ÊÓ

The substitution method is a powerful technique in calculus used to simplify integrals by transforming them into a form that is easier to evaluate. In this method, you replace a complicated part of the integrand with a single variable, making the integral more manageable.

In our example, the substitution used is:

• Letting \( u = x^{1/2} \), which turns \( x \) into \( u^2 \) and transforms the differential \( dx \) into \( 2u \, du \).

This substitution helps to convert the original integral:\[\int \frac{1}{x^{3/2} - \sqrt{x}} \, dx\]into a simpler integral:\[\int \frac{2}{u^2 - 1} \, du\]This approach effectively simplifies the expression by reducing complexities, such as roots or powers, into new variables. This can often reveal an underlying structure of the integral that is more straightforward to work with.Substitution is particularly useful when the integrand includes functions that, when substituted, simplify multiplication or make one factor a derivative of another.

Partial fraction decomposition is a technique used to break down complicated algebraic fractions into simpler ones, which can then be integrated individually. This is particularly useful for rational functions, where the numerator is of a lower degree than the denominator.

In our exercise:

• The integral simplifies to \( \frac{2}{u^2 - 1} \), where the denominator can be factored as \( (u-1)(u+1) \).

Once factorized, the next step is to express the fraction as the sum of simpler fractions:\[\frac{2}{(u-1)(u+1)} = \frac{A}{u-1} + \frac{B}{u+1}\]Solving for the constants \( A \) and \( B \), we find that:

• \( A = 1 \)
• \( B = 1 \)

This results in:\[\frac{2}{(u-1)(u+1)} = \frac{1}{u-1} + \frac{1}{u+1}\]Decomposition makes the integration process much simpler, as each term can now be integrated separately using basic integral rules. This technique is vital for handling integrals that involve rational expressions with polynomial denominators.

Logarithmic integration comes into play when integrating functions like \( \frac{1}{x} \), leading to results involving logarithms. This method is applied when dealing with partial fraction decompositions, as in the current example.

After decomposing into partial fractions, the integral:

• \( \int \left( \frac{1}{u-1} + \frac{1}{u+1} \right) \, du \)

can be resolved into:

• \( \ln|u-1| + \ln|u+1| + C \)

Here, the natural logarithm \( \ln \) is used because the derivative of \( \ln|x| \) is \( \frac{1}{x} \), making integration straightforward with log functions.

The property of logarithms, \( \ln|a| + \ln|b| = \ln|a \cdot b| \), allows us to combine these results:

• \( \ln|u^2 - 1| + C \)

Finally, substituting back \( u = x^{1/2} \) recreates the integral result in the original variable context. Applying logarithmic integration not only simplifies the process but allows handling of more complex integrals in a consistent manner.