Question

Perform the indicated integrations. $$ \int \frac{2 t^{2}}{2 t^{2}+1} d t $$

Short answer

The integral is \( t - \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t}{\sqrt{2}} \right) + C \).

Step-by-step solution

Identify the Integration Technique

When we have a rational function like \( \int \frac{2t^2}{2t^2 + 1} \, dt \), an effective approach is to consider polynomial long division if the numerator's degree is greater than or equal to the denominator's degree. Here, both are quadratic polynomials.

Simplify the Function

Divide the numerator by the denominator: \( \frac{2t^2}{2t^2 + 1} = 1 + \frac{-1}{2t^2 + 1} \). This simplification helps separate the integral into simpler parts.

Split the Integral

Rewrite the integral as two separate integrals: \( \int \left(1 + \frac{-1}{2t^2 + 1}\right) \, dt = \int 1 \, dt + \int \frac{-1}{2t^2 + 1} \, dt \). Each part can now be integrated individually.

Integrate the Constants

The integral of a constant \( 1 \) is simply \( t \), so \( \int 1 \, dt = t + C \).

Integrate the Rational Function

For the integral \( \int \frac{-1}{2t^2 + 1} \, dt \), use the formula \( \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \). Here, \( a = \sqrt{2} \), so \( \int \frac{-1}{2t^2 + 1} \, dt = -\frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t}{\sqrt{2}} \right) + C \).

Combine the Results

Add the results of both integrals: \( t - \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t}{\sqrt{2}} \right) + C \). This is the final integrated function.

Key concepts

Calculus

Calculus is a branch of mathematics that studies how things change. One of the main ideas of calculus is the concept of a limit, which leads to the definition of derivatives and integrals.

• Derivatives: Used to find the rate of change of a quantity.
• Integrals: Used to find the total accumulation of a quantity.

In our exercise, we focus on integration, which is the process of finding the integral of a function. Calculus provides various techniques to solve different types of integrals. For rational functions, which involve polynomials, specific techniques like polynomial long division can simplify the integration process. By understanding and applying these techniques, we can solve complex problems that arise in mathematical and real-world contexts.

Rational Functions

Rational functions are expressions formed by dividing two polynomials. They are important in calculus because they often appear in real-world applications and mathematical problems. A rational function has the form:\[ f(x) = \frac{P(x)}{Q(x)} \]where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) eq 0 \). In this exercise, we have a specific rational function:\[ \frac{2t^2}{2t^2 + 1} \]This expression presents an opportunity to use polynomial long division because the degrees of both the numerator and the denominator are equal. Understanding rational functions and how to manipulate them is crucial in simplifying the integration process. This allows more complex integrals to be broken down into manageable parts.

Polynomial Long Division

Polynomial long division is a method used to divide one polynomial by another, much like long division with numbers. This technique is especially helpful in calculus when dealing with rational functions.Here’s how polynomial long division works:

• Divide the first term of the numerator by the first term of the denominator.
• Multiply the entire divisor by this result and subtract it from the original polynomial.
• Repeat the process with the new polynomial formed after subtraction until reaching a remainder degree lower than the divisor.

In our exercise, dividing \( 2t^2 \) by \( 2t^2 + 1 \) results in a remainder of \( -1 \). This allows us to express the function as a simplified division plus a simpler fraction:\[ 1 + \frac{-1}{2t^2 + 1} \]This separation makes integrating each part individually more straightforward, showcasing the power of polynomial long division in calculus.

Integral Calculus

Integral calculus focuses on the concept of integration, which calculates the total accumulation of quantities. The integral can be thought of as finding the area under a curve or the total change over an interval.In our exercise, the integral involves two parts after simplification:1. The integral of a constant, \( \int 1 \, dt \), which equals \( t + C \).2. The integral of the rational function, \( \int \frac{-1}{2t^2 + 1} \, dt \), which uses the formula for integrating functions of the form \( \frac{1}{a^2 + x^2} \). This is linked to trigonometric functions, particularly the arctangent function. Using this formula, we find:\[ -\frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t}{\sqrt{2}} \right) + C \]Combining both results gives the final integral:\[ t - \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{t}{\sqrt{2}} \right) + C \]Understanding integral calculus is key to solving real-world problems where you need to calculate areas, volumes, and other quantities accumulated over time or space.