Question

Evaluate the following integrals: (i) \(\int \frac{x^{4}}{(1-x)^{3}} d x\) (ii) \(\int \frac{6 x^{2}-12 x+4}{x^{2}(x-2)^{2}} d x\)

Short answer

Based on the step-by-step solution provided: 1) The integral of \(\frac{x^4}{(1-x)^3} dx\) is: \(-[\frac{1}{2}(1-x)^{-2} + 2(1-x)^{-1} - 6\ln(1-x) - 2(1-x)^2 + \frac{1}{3}(1-x)^3] + C\) 2) The integral of \(\frac{6x^2 - 12x + 4}{x^2(x-2)^2} dx\) is: \(2\left[-\frac{1}{3}\ln(x) + \frac{1}{3x} + \frac{5}{3}\ln(x-2) - \frac{2}{x-2}\right] + C\)

Step-by-step solution

Simplify the integrand using substitution method

Let \(u = 1 - x\), so \(-du = dx\). Now, we have, \(x = 1-u\), and the integral becomes \(\int \frac{(1-u)^{4}}{u^{3}}(-du)\) Now, expand the numerator and integrate term by term.

Expand the numerator and integrate term by term

Expanding \((1-u)^4\) we get: \((1 - u)^4 = 1 - 4u + 6u^2 - 4u^3 + u^4\) Now, the integral is: \(- \int \frac{1 - 4u + 6u^2 - 4u^3 + u^4}{u^3}du = -\int u^{-3} - 4u^{-2} + 6u^{-1} - 4u + u^2 du\) Now, we can integrate term by term: \(- \int (u^{-3} - 4u^{-2} + 6u^{-1} - 4u + u^2) du\)

Integrate and substitute back

Integrating each term, we get: \(-[\frac{1}{2}u^{-2} + 2u^{-1} - 6\ln(u) - 2u^2 + \frac{1}{3}u^3] + C\) Now, substitute \(u = 1 - x\) back into the integrated expression: \(-[\frac{1}{2}(1-x)^{-2} + 2(1-x)^{-1} - 6\ln(1-x) - 2(1-x)^2 + \frac{1}{3}(1-x)^3] + C\) This is the final answer for the first integral. (ii) \(\int \frac{6x^{2} - 12x + 4}{x^{2}(x-2)^{2}} dx\)

Factor out common terms

We can factor out a 2 and simplify the integrand: \(\int \frac{6x^{2} - 12x + 4}{x^{2}(x-2)^{2}} dx = 2\int \frac{3x^{2} - 6x + 2}{x^{2}(x-2)^{2}} dx\)

Use partial fraction decomposition

Let's decompose the integrand into simpler fractions: \(\frac{3x^{2} - 6x + 2}{x^{2}(x-2)^{2}} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-2} + \frac{D}{(x-2)^2}\) Now, clear the fractions by multiplying through by \(x^2(x-2)^2\): \(3x^2 - 6x + 2 = A(x^2)(x-2)^2 + B(x)(x-2)^2 + C(x^2)(x)(x-2) + D(x^2)(x)\)

Determine constants

Solve for the constants A, B, C, and D using the system of linear equations: \(A = -\frac{1}{3}, \quad B = -\frac{1}{3}, \quad C = \frac{5}{3}, \quad D = 2\) Now, substitute these values back into the partial fraction decomposition: \(\frac{3x^{2} - 6x + 2}{x^{2}(x-2)^{2}} = -\frac{1}{3x} - \frac{1}{3x^2} + \frac{5}{3(x-2)} + \frac{2}{(x-2)^2}\)

Integrate the decomposed fractions

Now, integrate each term: \(2\int \left(-\frac{1}{3x} - \frac{1}{3x^2} + \frac{5}{3(x-2)} + \frac{2}{(x-2)^2}\right) dx\) \(2\left[-\frac{1}{3}\ln(x) + \frac{1}{3x} + \frac{5}{3}\ln(x-2) - \frac{2}{x-2}\right] + C\) This is the final answer for the second integral.

Key concepts

Integral Calculus

Integral calculus is a fundamental aspect of mathematics, primarily concerned with the accumulation of quantities and the area under and between curves. It allows for the calculation of a region's area, volume, or length when the geometry is too complex for basic geometrical formulas. In practice, integrating a function can be seen as determining the total amount of a quantity given its density at each point.

Integral calculus pairs with differential calculus, and together they make up the core of calculus, dealing with the behaviour of continuous functions. The main tool in integral calculus is the definite integral, which is used to calculate accumulations of a function over an interval. The indefinite integral, or antiderivative, represents a family of functions whose derivatives are the original function, and its calculation is fundamental to solving many mathematical problems.

U-Substitution

U-substitution, also known as the substitution rule, is one of the standard methods for performing integrals in calculus, analogously serving as the reverse chain rule of integration. The idea is to simplify a given integral by substituting part of the integrand with a new variable, usually denoted as 'u'. This can make an otherwise complex integral more manageable.

In the exercise, the substitution method simplifies the integrand from \(\int \frac{x^{4}}{(1-x)^{3}} dx\) to \(\int (u^{-3} - 4u^{-2} + 6u^{-1} - 4u + u^2) du\), a sum of powers of u that are more straightforward to integrate. The challenge usually lies in choosing the right substitution and managing differentials correctly.

Partial Fraction Decomposition

Partial fraction decomposition is an algebraic technique used to break down complex rational functions into simpler fractions that are easier to integrate. This method is particularly useful when dealing with integrands that are proper rational expressions, where the degree of the numerator is less than the degree of the denominator.

The goal is to represent the original function as a sum of simpler fractions, where each term corresponds to a factor of the denominator. After determining the coefficients of these simpler fractions, the problem of integrating the original function reduces to integrating several much easier functions. As in our exercise, applying partial fraction decomposition transformed the complex integrand into a sum of simple fractions, each of which can be integrated individually.

Integration Techniques

Various integration techniques are used to tackle complex integrals that cannot be handled via direct methods. These techniques include, but are not limited to, u-substitution, partial fraction decomposition, integration by parts, trigonometric substitution, and using tables of integrals.

Each method has its own cases where it's most effective. U-substitution is a versatile technique and often the first to be tried. Partial fraction decomposition is great for rational functions with polynomial denominators. Integration by parts helps with products of functions. Trigonometric substitution simplifies integrals containing square roots of quadratic expressions, and tables of integrals provide quick results for known integral forms. To master integral calculus, a thorough understanding of each technique and its applications is crucial.