91Ó°ÊÓ

In integral calculus, the substitution method is an essential technique, often used to simplify integrals and make them easier to evaluate. The substitution method involves changing the variable of integration to transform the integral into a more standard form, making it easier to solve.

When using this method, it’s important to:

• Identify the part of the integrand that can be substituted.
• Determine a new variable, say \(u\), and express this in terms of the original variable.
• Derive \(du\) by differentiating \(u\) with respect to the original variable.
• Rewrite the entire integral in terms of \(u\).
• Solve the new integral, which should be in a simpler form.

In this exercise, \(u = z^2\) is chosen as the substitution, simplifying the integral by transforming the original expression \(\int \frac{z}{9-z^4} \, dz\) into a simpler form, \(\frac{1}{2} \int \frac{1}{9 - u^2} \, du\). This process helps to reduce the original complexity of the integral.

Integral tables are handy tools in calculus that list various standard integrals and their solutions. These tables provide quick references for integrals of different forms, which have been pre-solved for ease of use. When working on integral calculus problems, checking an integral table can save time and guide the solving process.

To use an integral table effectively, one should:

• Identify the form of the integrand in your given integral.
• Match this form to a similar expression in the table.
• Apply the solution provided in the table, modifying constants as necessary.

In the problem outlined, after substitution, the integral \(\frac{1}{2} \int \frac{1}{9-u^2} \, du\) closely matches a standard form found in integral tables involving logarithmic functions, guiding us to the correct solution involving inverse hyperbolic functions.

Inverse hyperbolic functions and their integrals appear often in calculus, much like their trigonometric analogs. These functions include \(\sinh^{-1}\), \(\cosh^{-1}\), and \(\tanh^{-1}\). They are used in integration when expressions involve squares and square roots, similar to trigonometric relationships.

In our solution, the result relates to a logarithmic representation tied to hyperbolic functions. Often, integrals such as \(\int \frac{1}{a^2 - u^2} \, du\) can be solved using inverse hyperbolic functions, yielding forms like:

• Hyperbolic sine, \(\sinh^{-1}(x)\): \(\int \frac{1}{\sqrt{a^2 + u^2}} \, du\) yields a result like \(\sinh^{-1}(\frac{u}{a})\).
• Hyperbolic tangent, \(\tanh^{-1}(x)\): \(\int \frac{1}{a^2 - u^2} \, du\) results in logarithmic expressions similar to inverse trigonometric results.

For the given integral, this leads to a logarithmic solution, \(\frac{1}{6} \ln \left| \frac{3 + z^2}{3 - z^2} \right| + C\), which is typical for integrations involving inverse hyperbolic relationships.