Question

In Exercises \(5-14,\) evaluate the integral. $$\int \frac{7 d x}{2 x^{2}-5 x-3}$$

Short answer

The integral is equal to \( \frac{21}{4} \ln |2x + 1| - \frac{7}{2} \ln |x - 3|+ C \), where C is the constant of integration.

Step-by-step solution

Factorize the denominator

We first factorize \(2 x^{2}-5 x-3=0\). By looking for two numbers that add up to -5 and multiply to -6, we can factorize the quadratic as \(2x^{2}-5x-3 = (2x+1)(x-3)\)

Apply partial fraction decomposition

Apply partial fractions to decompose the integral. We write \(\frac{7}{2x^{2}-5x-3} = \frac{A}{2x+1} + \frac{B}{x-3}\), where A and B are constants that we need to solve for.

Solve for A and B

To solve for A and B, we multiply both sides by \(2x^2-5x-3\) to get rid of the denominators. We obtain: \(7 = A . (x - 3) + B . (2x + 1)\). Solve this equation by substituting appropriate x-values that will lead to one of A or B. From factorisation, we know that if \(x = 3\) or\(x = -\frac{1}{2}\), the equation will become zero. Plug these values into the equation and solve for A and B.

Substitute A and B back into the integral and solve

Substitute the values of A and B back into the integral and break it down into two separate integrals: \(\int \frac{7}{2 x^{2}-5 x-3} d x=\int \frac{A}{2x+1} d x + \int \frac{B}{x-3} d x\). These integrals can be solved using basic integration rules.

Key concepts

Integration Techniques

In calculus, integration is fundamental, and mastering various integration techniques is crucial for solving complex integrals. When faced with an algebraic expression under an integral, one such technique that often comes to the rescue is partial fraction decomposition. This method involves breaking down a complex rational expression into simpler fractions that can be integrated individually.

Consider the integral \( \int \frac{7 dx}{2x^2-5x-3} \). It's not immediately obvious how to integrate this directly; however, by applying partial fraction decomposition, the process becomes more manageable. Once the denominator is factorized, the integral is expressed as the sum of simpler fractions, and basic integration rules can be applied to these fractions. It's akin to turning a daunting task into smaller, more approachable subtasks that can be handled with ease.

Factorization of Polynomials

The success of partial fraction decomposition often hinges on the ability to factorize polynomials. Factorization is the process of breaking down a polynomial into a product of its factors, which are usually of lower degree and simpler to deal with. In the given exercise, we are presented with the quadratic polynomial \(2x^2-5x-3\).

The goal is to express it as a product of two first-degree polynomials. By finding two numbers that simultaneously add to the middle coefficient, -5, and multiply to the constant term multiplied by the leading coefficient, -6, we identify the factors \(2x+1\) and \(x-3\). This step is vital because, without it, partial fraction decomposition wouldn't be possible. It's crucial to view factorization as a puzzle where the pieces need to fit together perfectly to unlock the next steps towards solving the integral.

Definite Integral Calculation

The problem and solution provided do not involve a definite integral calculation, nonetheless, understanding how to approach such calculations is an important skill. A definite integral represents the area under the curve of a function between two bounds. To evaluate a definite integral, one would generally find the antiderivative of the function and then apply the boundaries.

In a scenario where partial fraction decomposition is used within the context of a definite integral, each simpler fraction would be integrated separately over the given interval. The key is to remain methodical: after integrating, carefully substitute the upper and lower limits into the antiderivatives, and then calculate the difference. This meticulous approach can reveal valuable aspects of functions and their graphs, enhancing one's overall comprehension of calculus.