Question

Find: (a) \(\int(x+3)(x+1)^{1 / 2} d x\) (b) \(\int x \operatorname{tr} x d x \quad(x>0)\)

Short answer

(a) \(\frac{2}{5}(x+1)^{5/2} + \frac{4}{3}(x+1)^{3/2} + C\); (b) \(\frac{x^2}{2} \ln x - \frac{x^2}{4} + C\).

Step-by-step solution

Identify Integration Method for Part (a)

For the integral \( \int(x+3)(x+1)^{1/2} \, dx \), we recognize that the integrand has a term of the form \((x+1)^{1/2}\). This suggests using substitution. Let \(u = x+1\), then \(du = dx\) and \(x = u-1\). This changes the integral to \(\int(u-1+3)u^{1/2} \, du\).

Simplify and Integrate Part (a)

Substitute the identity \(u = x+1\) into the integral: \(\int (u+2)u^{1/2} \, du = \int u^{3/2} + 2u^{1/2} \, du\). Now integrate term by term: \(\int u^{3/2} \, du = \frac{2}{5}u^{5/2}\) and \(\int 2u^{1/2} \, du = 2 \times \frac{2}{3}u^{3/2} = \frac{4}{3}u^{3/2}\).

Evaluate and Substitute Back in Part (a)

The integrals from the previous step are \(\frac{2}{5}u^{5/2} + \frac{4}{3}u^{3/2}\). Re-substitute \(u = x+1\) back into the expression to obtain: \(\frac{2}{5}(x+1)^{5/2} + \frac{4}{3}(x+1)^{3/2} + C\), where \(C\) is the constant of integration.

Identify Integration Method for Part (b)

For the integral \(\int x \operatorname{tr} x \, dx\), note that \( \operatorname{tr} \) denotes the trace of a matrix, but here it likely refers to a function involving \(x\). Assume a standard integral form for simplicity. Assuming \( \operatorname{tr} x = \ln x \) given the context, we have \(\int x \ln x \, dx\). To solve, use integration by parts where \(u = \ln x\) and \(dv = x \, dx\).

Perform Integration by Parts for Part (b)

Apply integration by parts formula, \(\int u \, dv = uv - \int v \, du\). Let \(u = \ln x\) and \(dv = x \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^2}{2}\). Substitute into the formula to get \[\frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln x - \int \frac{x}{2} \, dx\].

Integrate Remaining Part and Simplify Part (b)

Simplify \(\int \frac{x}{2} \, dx\) to \(\frac{1}{2} \int x \, dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4}\). Substitute back: \(\frac{x^2}{2} \ln x - \frac{x^2}{4} + C\), where \(C\) is the constant of integration.

Key concepts

Integration by Substitution

Integration by substitution is a powerful technique often used to simplify complex integrals. When faced with an integral like \( \int(x+3)(x+1)^{1/2} \, dx \), the complication arises from the combination of terms involving a function and its derivative. This technique involves substituting a part of the integral with a new variable, thus transforming it into a simpler one. Let’s explore how it works.

In this example, we notice a term \((x+1)^{1/2}\), a perfect candidate for substitution. We set \(u = x+1\), which simplifies our task because the derivative \(du = dx\). This means the differential stays straightforward. Also, since \(x = u-1\), we can reformulate the integral in terms of \(u\).

This substitution process transforms our original integral into \(\int (u-1+3)u^{1/2} \, du = \int u^{3/2} + 2u^{1/2} \, du\). By eliminating the more complicated expressions and dealing with powers of \(u\), we make the problem more approachable.

Once we've integrated, the last step involves substituting back the original variable to express the solution in terms of \(x\). This method is not only about transforming variables, but also about making the integral much more manageable by focusing on simplifying complex expressions.

Integration by Parts

Integration by parts is a valuable method for tackling integrals involving a product of two functions. The formula we use is derived from the product rule for differentiation and is expressed as \(\int u \, dv = uv - \int v \, du\). Understanding and identifying which functions to assign to \(u\) and \(dv\) are crucial for successful application.

Let’s consider the integral \(\int x \ln x \, dx\). Here, we assume the role of \(\operatorname{tr} x\) is similar to the function \(\ln x\). By taking \(u = \ln x\) and \(dv = x \, dx\), we effectively choose \(u\) to be a function that decreases in complexity upon differentiation. Computing \(du\) gives us \(\frac{1}{x} \, dx\) while integrating \(dv\) results in \(v = \frac{x^2}{2}\).

Substituting these into the formula, \( \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln x - \int \frac{x}{2} \, dx \), we can further simplify to obtain \(\frac{x^2}{2} \ln x - \frac{x^2}{4} + C\). Through choosing suitable \(u\) and \(dv\), integration by parts effectively addresses daunting products of functions.

Always think about how each choice affects the complexity of the integral you're trying to solve. It’s about finding a balance that leads to a simpler expression after computing the derivative and the integral.

Definite and Indefinite Integrals

Definite and indefinite integrals are foundational concepts in calculus that deal with the process of integration. Understanding these two types is crucial for solving a wide range of mathematical problems.

Indefinite integrals are what you encounter when there are no limits of integration specified, represented as \(\int f(x) \, dx\). The result of an indefinite integral is a family of functions, denoted by the inclusion of a constant of integration, \(C\). This constant represents the unknown shift of the antiderivative on the y-axis and is essential for capturing all possible solutions.

In our examples, both \(\int (x+3)(x+1)^{1/2} \, dx\) and \(\int x \ln x \, dx\) are indefinite integrals. Their solutions, \(\frac{2}{5}(x+1)^{5/2} + \frac{4}{3}(x+1)^{3/2} + C\) and \(\frac{x^2}{2} \ln x - \frac{x^2}{4} + C\), include this constant, reflecting the indefinite nature of the integrals.

Definite integrals, on the other hand, involve specific bounds and calculate the net area under the curve between these limits. They are expressed as \(\int_{a}^{b} f(x) \, dx\), providing a numerical result. While this article focuses on indefinite integrals, understanding how both definite and indefinite integrals complement each other enriches your grasp of calculus's versatility.