Question

Evaluate the following integrals. $$\int \frac{\ln ^{2} x+2 \ln x-1}{x} d x$$

Short answer

The final result of the integral is: $$\int \frac{\ln^2 x + 2\ln x -1}{x} dx = (\ln^2 x)(\ln x) +2 (\ln^2 x) - (\ln x) + C$$ Where C is the constant of integration.

Step-by-step solution

Apply integration by parts for the first term

Recall the integration by parts formula: $$\int u dv = uv - \int v du$$ Let's apply integration by parts for the first term, where: $$u = \ln^2 x$$ $$dv = \frac{1}{x} dx$$ To find du, we differentiate u with respect to x: $$du = 2(\ln x) \cdot \frac{1}{x} dx$$ Now, we integrate dv: $$v = \int \frac{1}{x} dx = \ln x$$ Now, using the integration by parts formula for the first term, we get: $$\int \frac{\ln^2 x}{x} dx = (\ln^2 x)(\ln x) - \int (\ln x)(2 \cdot \frac{\ln x}{x}) dx$$

Integrate the remaining terms and rewrite

We can integrate the second term and the last term of the integral as follows: $$\int \frac{2 \ln x}{x} dx = 2\int \frac{\ln x}{x} dx$$ And, $$\int -\frac{1}{x} dx = -\int \frac{1}{x} dx$$ Now, combining all terms, we have: $$\int \frac{\ln^2 x + 2\ln x -1}{x} dx = (\ln^2 x)(\ln x) - \int 2(\frac{\ln x }{x})^2 dx +2\int \frac{\ln x}{x} dx - \int \frac{1}{x} dx$$

Simplify and apply integration by parts again

The first integral on the right side can be simplified further. Let's apply integration by parts again to the second term, where: $$u = \ln x$$ $$dv = -2 \cdot \frac{\ln x}{x} dx$$ To find du, we differentiate u with respect to x: $$du = \frac{1}{x} dx$$ Now, we integrate dv: $$v = \int -2 \cdot \frac{\ln x}{x} dx = -2 (\ln x)$$ Now, using the integration by parts formula, we get: $$-\int 2(\frac{\ln x }{x})^2 dx = (\ln x)(-2 (\ln x)) - \int (-2(\ln x)) \frac{1}{x} dx$$ $$= -2 (\ln^2 x) + 2\int \frac{\ln x}{x} dx$$ Now, replace this into the expression from Step 2: $$\int \frac{\ln^2 x + 2\ln x -1}{x} dx = (\ln^2 x)(\ln x) - (-2 (\ln^2 x) + 2\int \frac{\ln x}{x} dx) +2\int \frac{\ln x}{x} dx - \int \frac{1}{x} dx$$

Final simplification

Simplify and combine like terms: $$\int \frac{\ln^2 x + 2\ln x -1}{x} dx = (\ln^2 x)(\ln x) +2 (\ln^2 x) - \int \frac{1}{x} dx$$ Now, integrate the last term: $$\int \frac{\ln^2 x + 2\ln x -1}{x} dx = (\ln^2 x)(\ln x) +2 (\ln^2 x) - (\ln x) + C$$ Where C is the constant of integration.

Key concepts

Logarithmic Functions

Logarithmic functions, such as the natural logarithm \((\ln x)\), are fundamental in calculus due to their unique properties. These functions help in expressing exponential growth or decay in mathematical models. When you see \(\ln x\), it represents the power to which the base of the natural log, \(e \approx 2.718\), must be raised to obtain \(x\).
Logarithmic differentiation and integration are important because they simplify complex problems. In our original exercise, we encounter logarithmic functions squared (\(\ln^2 x\)), which adds another layer of complexity. However, with consistent practice of handling these functions through basic rules of differentiation and integration, you'll find these tasks become more manageable.
Here are some key points about logarithmic functions that can ease their integration:

• The derivative of \((\ln x)\) is \((1/x)\).
• Functions like \((\ln^2 x)\) require you to use product or chain rules during differentiation.
• Integration can be approached using techniques like integration by parts as seen in the exercise.

Understanding the properties of logarithmic functions helps in unlocking more advanced calculus concepts and solutions.

Definite and Indefinite Integrals

Integrals are the building blocks for understanding areas under curves and accumulation of quantities. When working with integrals, it's essential to distinguish between definite and indefinite types.
Indefinite integrals are the most common type you'll see. These are integrals without specified limits of integration and include a constant of integration, \(C\), because the process of integration can yield multiple functions differing by a constant.
Definite integrals, on the other hand, come with upper and lower limits indicating the interval across which the function is to be integrated. These provide exact numerical values, representing area under the curve within the specified limits.

• Definite integrals provide an accumulated quantity like area, while indefinite integrals provide the antiderivative.
• In our exercise, the integration is indefinite, highlighted by the inclusion of the constant \(C\).
• The integration techniques, whether definite or indefinite, require a systematic approach, as illustrated in multiple steps for our problem.

Grasping this distinction is crucial as it affects how you interpret and solve integration problems, helping to refine numerical or functional outcomes accordingly.

Integration Techniques

Integration techniques such as integration by parts are essential tools in calculus. These techniques can simplify complex integrals that are not solvable using standard rules like the power rule or the basic antiderivatives.
Integration by parts arises from the product rule of differentiation and is used to integrate products of functions. The formula is: \(\int u \ dv = uv - \int v \ du \) where you choose parts of the integrand to be \(u\) and \(dv\). Choosing wisely can simplify the problem significantly.
In the original exercise, integration by parts was used to break down the function featuring \(\ln^2 x\). Here’s how you can effectively use integration by parts:

• Select \(u\) such that its derivative, \(du\), simplifies the remaining integral.
• \(dv\), the differential part, should be easily integratable to find \(v\).
• Apply repeatedly if the resulting integral remains complex, as seen by applying integration by parts twice.

Mastering these techniques allows you to approach a wider range of integrals with confidence and efficiency. It's a skill gained through practice and strategic thinking in solving mathematical problems.