Question

Evaluate the following integrals: (i) \(\int \frac{x^{3}+x^{2}+x+3}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d\) ) (ii) \(\int \frac{d x}{x^{4}\left(x^{3}+1\right)^{2}}\) (iii) \(\int \frac{x^{7}+2}{\left(x^{2}+x+1\right)^{2}} d x\) (iv) \(\int \frac{3 x^{4}+4}{x^{2}\left(x^{2}+1\right)^{3}} d x\)

Short answer

Question: Evaluate the integral \(\int \frac{x^{3}+x^{2}+x+3}{\left(x^{2}+1\right)\left(x^{2}+3\right)} dx\). Answer: \(2 \arctan(x) + \frac{1}{2} \ln |x^2 + 3| + C\)

Step-by-step solution

Partial Fraction Decomposition

For the first integral, we need to decompose the given function into the sum of partial fractions. So, we have to find constants A, B, C, and D such that: \(\frac{x^{3}+x^{2}+x+3}{\left(x^{2}+1\right)\left(x^{2}+3\right)} = \frac{A x + B}{x^{2} + 1} + \frac{C x + D}{x^{2} + 3}\) Now, we need to find A, B, C, and D.

Find Constants A, B, C, and D

To find A, B, C, and D, we'll clear the denominators and combine the terms: \(x^3 + x^2 + x + 3 = A x (x^2 + 3) + (Cx + D)(x^2 + 1)\) Expanding and collecting like terms, we get: \(x^3 + x^2 + x + 3 = (A + C) x^3 + (A + D)x^2 + (3A+B+ C)x + 3D\) Comparing the coefficients, we have the following system of equations: \(A + C = 1\) \(A + D = 1\) \(3A + B + C = 1\) \(3D = 3\) Solving this system of equations, we get A = 0, B = 2, C = 1, and D = 1. So the decomposition becomes: \(\frac{2}{x^2 + 1} + \frac{x + 1}{x^2 + 3}\)

Evaluate the Integral

Now we can rewrite the integral as: \(\int \frac{x^{3}+x^{2}+x+3}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d = \int \frac{2}{x^2 + 1} d + \int \frac{x + 1}{x^2 + 3} d\) The first integral can be easily evaluated as: \(\int \frac{2}{x^2 + 1}dx = 2\int \frac{1}{x^2 + 1}dx = 2 \arctan(x) + C_1\) For the second integral, perform a substitution with \(x^2 + 3 = u\) and \(2x dx = du\): \(\int \frac{x + 1}{x^2 + 3} dx = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln |u| + C_2 = \frac{1}{2} \ln |x^2 + 3| + C_2\) Combining the two integrals, our final answer is: \(2 \arctan(x) + \frac{1}{2} \ln |x^2 + 3| + C\) This is our solution for the integral (i). Likewise, we'll work on the integrals (ii), (iii), and (iv) in a step-by-step approach, decomposing when necessary and applying different techniques as needed.

Key concepts

Partial Fraction Decomposition

Understanding partial fraction decomposition can be crucial when tackling complex rational functions in integral calculus. It's like unraveling a tightly knit sweater into strands of yarn that are much easier to work with. This technique involves breaking down a complex fraction into simpler 'partial' fractions that are more manageable to integrate.

For example, consider a fraction with a polynomial numerator and a denominator that can be factored into simpler polynomials. We propose an equation where the original fraction equals the sum of several fractions, each with its separate simpler denominator and an unknown numerator (usually a constant for linear factors or a linear polynomial for quadratic factors).

Once you've set up your equation, the goal is to find the values of these unknown numerators that will satisfy the equation. This involves multiplying through by the common denominator to eliminate the fractions, thus obtaining an equation in terms of polynomials. Equating coefficients of like terms on both sides gives us a system of equations that, when solved, provide the coefficients for our partial fractions. The result is a more straightforward expression that often leads to simpler integrals.

Integration Techniques

There's a whole toolbox of integration techniques used in calculus, each method suited for a different type of function or integral form. Among them, substitution is widely used, especially when an integral contains a function and its derivative. It simplifies the integral, turning it into a basic form we can integrate easily.

Another powerful method is integration by parts, which is essential when the integrand is the product of two functions. The technique involves choosing one function to differentiate and another to integrate, which can make the integral less daunting. Additionally, for complex functions that are the product of a polynomial and a transcedental function, the method of polynomial long division can be employed before integration.

Finally, trigonometric substitution comes in handy when dealing with integrals containing square roots of quadratic expressions. It transforms the integral into a trigonometric identity. When facing a challenging integral, choosing the right technique can mean the difference between an unsolvable problem and a neat solution.

Indefinite Integrals

Delving into the world of indefinite integrals, you're essentially looking for the antiderivative of a function, which represents the family of all functions whose derivative is the original function. Unlike definite integrals, indefinite integrals don't have bounds and are, hence, associated with a constant of integration to account for all possible solutions.

They're typically denoted by the integral symbol followed by the function and the differential, such as \( \int f(x) \, dx \). The process of integration, just like differentiation, has its own set of rules and formulas. For example, the power rule of integration allows us to integrate powers of x, while the sum rule lets us integrate a sum of functions term by term.

One important advice when dealing with indefinite integrals is to carefully check and simplify the function before attempting to integrate. This may involve expanding expressions, factoring, or even recognizing when a function's appearance can be deceivingly complex but is indeed a simple derivative in disguise.