Question

$$\text {Evaluate the following integrals.}$$ $$\int \frac{x^{3}+2}{x^{3}-2 x^{2}+x} d x$$

Short answer

Question: Evaluate the integral $$\int \frac{x^{3}+2}{x^{3}-2x^{2}+x} dx$$ Answer: $$2\ln|x| - \ln|x-1| - \frac{3}{x-1} + C$$

Step-by-step solution

Partial Fraction Decomposition

In this step, we will break down the given rational function into simpler fractions using partial fraction decomposition. To do this, we first need to factor the denominator: $$x^{3} - 2x^{2} + x = x(x^{2} - 2x + 1) = x(x-1)^2$$ Now we can decompose the given integrand: $$\frac{x^{3}+2}{x(x-1)^2} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$$ Multiplying both sides by the denominator \( x(x-1)^2 \), we get: $$x^{3}+2 = A(x-1)^2 + Bx(x-1) + Cx$$ Now we need to find the values of A, B, and C.

Solve for A, B, and C

We can find the values of A, B, and C by substituting suitable x values to simplify the equation: 1. Set x=0: $$ 2 = A(1)^2 \Rightarrow A = 2 $$ 2. Set x=1: $$ 1 + 2 = C(1) \Rightarrow C = 3 $$ 3. Now, we need to find B. Let's expand the given equation and collect terms: $$x^{3}+2 = 2(x-1)^2 + Bx(x-1) + 3x$$ $$x^{3}+2 = 2(x^{2}-2x+1) + Bx^{2}-Bx^{3} + 3x$$ Now, comparing the powers of x, we get: $$ -B = 1 (coefficients\ of\ x^{3})$$ $$ B = -1 $$ Now that we have found A, B, and C, we can rewrite our decomposed integrand: $$\frac{x^{3}+2}{x(x-1)^2} = \frac{2}{x} - \frac{1}{x-1} + \frac{3}{(x-1)^2}$$

Integrate the decomposed integrand

Now, we can integrate the decomposed integrand term by term: $$\int \frac{x^{3}+2}{x(x-1)^2} dx = \int \frac{2}{x} dx - \int \frac{1}{x-1} dx + \int \frac{3}{(x-1)^2} dx$$ Using the power rule for integration and substitution for each term: $$ 2\int \frac{1}{x} dx - \int \frac{1}{x-1} dx + 3\int \frac{1}{(x-1)^2} dx$$ $$ = 2(\ln|x|) - (\ln|x-1|) - \frac{3}{x-1} + C$$ Our final antiderivative is: $$2\ln|x| - \ln|x-1| - \frac{3}{x-1} + C$$

Key concepts

Integration Techniques

Integration is a fundamental technique in calculus, primarily used to find areas, volumes, and the total accumulation of a quantity. When it comes to integrating rational functions, i.e., functions that are the ratio of two polynomials, one effective method is partial fraction decomposition. This technique breaks down complex fractions into simpler, more manageable ones, often resulting in elementary functions we already know how to integrate.

For instance, in the given exercise, the fraction \(\frac{x^{3}+2}{x^{3}-2x^{2}+x}\) is decomposed into a sum of simpler fractions, which are easier to integrate individually. By finding constants that satisfy the equivalent expression, each term becomes trivial to integrate using basic integration rules or substitutions. This step-by-step method not only simplifies the process of finding the antiderivative but also opens doors to integrating more complex rational functions systematically and reliably.

Rational Functions

A rational function is formed when one polynomial is divided by another. It's represented as \(\frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x)\) is not equal to zero. The rich variety of their behavior near points where the denominator equals zero (poles) or the numerator and denominator share common factors (zeroes) is what makes integrating rational functions quite intriguing.

The complexities of rational functions are what typically lead to the use of partial fraction decomposition in integration. By breaking down the original function into partial fractions, we can work with simpler expressions that are easier to integrate. During the process, a thorough understanding of polynomial division, factoring, and simplifying algebraic expressions is crucial. This allows the creation of a sum of simpler fractions that are much more approachable, as demonstrated in the exercise. The ability to manipulate these algebraic structures is fundamental in problem-solving within calculus.

Antiderivatives

The antiderivative of a function, often represented with a capital \(F\) if \(f\) is the original function, is a function whose derivative is the original function. In other words, \(F' = f\). The antiderivative, also known as the indefinite integral, represents the accumulation of a quantity and is essential in various areas such as physics for calculating position from a velocity function.

When finding antiderivatives, we use various techniques, and one important rule is the power rule for integration, which was applied to each term after decomposition in the given exercise. The power rule states that the integral of \(x^n\) with respect to x is \(\frac{1}{n+1}x^{n+1}\), provided \(n eq -1\). For \(n = -1\), which corresponds to \(\frac{1}{x}\), the antiderivative is the natural logarithm of the absolute value of x. In the solution to our exercise, we can see this application clearly, which simplifies the integration process greatly and brings us a step closer to finding the accumulated quantity described by the original function.