Question

Evaluate the following integrals. Assume a and b are real numbers and \(n\) is a positive integer. \(\int \frac{x}{a x+b} d x\) (Hint: \,\(u=a x+b\).)

Short answer

Question: Evaluate the integral \(\int \frac{x}{ax+b} dx\) using the given substitution \(u = ax + b\). Answer: \(\int \frac{x}{a x+b} d x = \frac{1}{a}(ax+b - b \ln{ |ax + b| }) + C\).

Step-by-step solution

Apply the substitution

Let \(u = ax + b\). Now find the derivative of \(u\) with respect to \(x\): \(\frac{d u}{d x} = a\).

Express the integral in terms of \(u\)

Replace \(x\) within the integrand with the expression in terms of \(u\). Also, substitute the derivative \(d u\) in place of \(dx\). The integral becomes, \(\int \frac{x}{u} \frac{d u}{a}\).

Simplify the integral

Simplify the integrand by canceling out the terms: \(\int \frac{x}{u} \frac{d u}{a} = \frac{1}{a} \int \frac{x}{u} d u\). Now, replace \(x\) using the expression \(u = ax + b\): \(\frac{1}{a} \int \frac{u - b}{u} d u\).

Separate the integrand into two fractions

Rewrite the integrand as the sum of two fractions: \(\frac{1}{a} \int (\frac{u}{u} - \frac{b}{u}) d u\).

Evaluate the integral of each term

Perform the integral for each term: \(\frac{1}{a} \int 1 - \frac{b}{u} d u = \frac{1}{a} (u - b \ln{ |u| }) + C\).

Convert back to \(x\)

Recall that \(u = ax + b\), and substitute this value back into the result: \(\frac{1}{a}(ax+b - b \ln{ |ax + b| }) + C\). Finally, the evaluated integral is: \(\int \frac{x}{a x+b} d x = \frac{1}{a}(ax+b - b \ln{ |ax + b| }) + C\).

Key concepts

Definite Integrals

Definite integrals are a way to calculate the area under a curve, from one specific point to another, on a given function. They are expressed in the form \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the limits of integration. These limits indicate the interval over which you find the area.
Definite integrals require no arbitrary constant \(C\) because the calculation provides a fixed numerical value. When solving definite integrals, the goal is often to substitute to simplify the function, then calculate the area by finding the anti-derivative and evaluating it at the upper and lower limits. Remember:

• Use substitution to make the integrand simpler.
• Find the anti-derivative.
• Evaluate the difference between the upper and lower limits of the definite integral.
• The result gives the area under the curve between the two limits.

This process requires careful work to ensure accuracy in the simplification and proper application of limits.

Indefinite Integrals

Indefinite integrals are the general form of integration and involve finding the antiderivative of a given function. They differ from definite integrals because they do not have limits of integration.
The result of an indefinite integral is a function plus a constant, \( C \), expressed as \(\int f(x) \, dx = F(x) + C\). This constant appears because the process of differentiation can lose information about constant values.To solve an indefinite integral, you will usually:

• Determine a substitution if needed, which simplifies the integrand (as seen in our example with \(u = ax + b\)).
• Find the antiderivative of the integrand in terms of the substituted variable.
• Simplify and substitute back in terms of the original variable to get the final result.

Indefinite integrals are essential in providing the 'family' of functions whose derivative is the original function, allowing for flexible application with added constants.

Integral Calculus

Integral calculus is one of the two major branches of calculus, with the other being differential calculus. Integral calculus focuses on the concept of integrals, by dealing with the accumulation of quantities such as areas under a curve.
The key concepts of integral calculus include:

• Definite Integrals, which find the exact area under a curve between two limits.
• Indefinite Integrals, which are more general and involve finding antiderivatives without specific bounds.
• The Fundamental Theorem of Calculus, connecting differentiation and integration and showing that they are inverse processes.

In integral calculus, techniques like substitution help in evaluating integrals by simplifying complex functions into simpler elements. It relies heavily on recognizing patterns and understanding functions deeply to apply suitable methods for solving integration problems. Overall, integral calculus allows for understanding changes and total quantities effectively, linking geometric and analytical insights.