Problem 9 - Calculus of a Single Variable

Question

Finding an Indefinite Integral In Exercises \(1-20\) , find the indefinite integral. $$ \int \frac{t}{t^{4}+25} d t $$

Step-by-Step Solution

Answer

To solve this problem, we will use substitution method and factoring of polynomial to simplify the equation and then we will use standard integrating formulas to solve the integral. Substitution is the very crucial step for solving this problem.

Step 1: Identify the substitution

Let's designate \(u = t^{4}+25\). So, our integral will become \(\int \frac{t}{u}\). But we've to replace \(dt\) as well. We can do that by deriving our substitution equation in respect to \(t\). So \(du = 4t^{3} dt\). Now we have a \(t\) in our integral, so we need to solve our \(du\) in terms of \(t dt\). It will be \( t dt = \frac{du}{4t^{3}}\).

Step 2: Substitute and Simplify

Put these all equations to the original integral, we have: \[ \int \frac{1}{u} * \frac{du}{4t^{2}} \] which simplifies to \[ \frac{1}{4}\int \frac{du}{u * t^{2}} \] we can see \(t^{2}\) in denominator, for this we can substitute \(t^{2} = u - 25\). So, it becomes \[ \frac{1}{4}\int \frac{1}{u^{2} - 25} du \] Now this is an integral form which can be solved by using quadratic formula.

Step 3: Solve the Integral

Here, to solve this integral, we can first simplify the denominator by factoring to \((u - 5)(u+5)\). You will end up with simple integral which can be solved easily through known formulas and once it is calculated, replace \(u\) with \(t^{4}+25\) to get the result in terms of \(t\) which is the original variable.

Key Concepts

Substitution Method

The substitution method is a powerful technique in calculus, especially for solving integrals. It involves changing the original variable to a new variable that simplifies the integral. In the exercise given, we started by setting a new variable, \( u \), equal to \( t^4 + 25 \). This step changes the complex expression under the integral into a simpler form.
This substitution helps in rewriting the differential part \( dt \) in terms of \( du \), where \( dt \) is the differentiation of our initial substitute equation. In this case, calculating \( du \) provides \( du = 4t^3 \, dt \), allowing us to link \( t \, dt \) to \( du \) and seamlessly transition our problem from one variable to another. This step-by-step variable change not only simplifies the given integral but also sets the stage for easier integration techniques.

Integration Techniques

Integration techniques are various strategies used to calculate integrals, whether definite or indefinite. In our exercise, we applied substitution to switch variables, then needed to address the resulting expression. After substitution, we were left with an altered integral: \( \int \frac{1}{u \, t^2} \, du \), which still needed further adjustment.
Adjusting techniques are guided by simplifying the expression. In this case, recognizing that \( t^2 = u - 25 \) allows for an even more simplified integral. This showcases how different techniques can work together to gradually transform a complicated form into a solvable one. Each step leverages mathematical insights to reduce complexity, making the solution pathways clearer and more manageable.

Factorization

Factorization simplifies expressions, particularly polynomials, by breaking them into a product of simpler components. This method becomes particularly useful when dealing with quadratic forms in integrals. In our example, transforming the integral’s denominator \( u^2 - 25 \) involves identifying it as a difference of squares.
The difference of squares, \( u^2 - 25 \), presents itself algebraically as \( (u - 5)(u + 5) \). Recognizing this pattern simplifies the integral significantly, allowing us to progress with simpler fractions that are easier to integrate. This intuition often requires practice to spot, but mastering it is crucial for addressing a range of integration and differential equations.

Quadratic Formula

The quadratic formula, typically used to solve quadratic equations, makes an appearance in this integral when considering integrals involving squared terms. Though not directly used in the solution calculation, the concepts of quadratic adjustments guide simplification attempts.
In some integrals, the quadratic formula provides a way to decompose or rearrange terms, aiding in integration. Understanding it fully involves recognizing standard quadratic scenarios and how they transform expressions. While factorization resolved our integral here, the quadratic formula remains a vital tool across many calculus problems, providing structured preprocessing for many integrals involving quadratic expressions.