91Ó°ÊÓ

Integration by parts is a method often used to integrate products of functions. You can think of it as the reverse of the product rule for differentiation. In this problem, we need to find the integral of the natural logarithm function, which isn't directly integrable by common rules.
Instead, we apply the integration by parts formula, given by: \[\int u \, dv = uv - \int v \, du\]
For our integral \( \int \ln(x+2) \, dx \), we let \( u = \ln(x+2) \) and \( dv = dx \). - Differentiating \( u \), we get \( du = \frac{1}{x+2} \, dx \).- Integrating \( dv \), we obtain \( v = x \).

Substituting back into the integration by parts formula gives:\[\int \ln(x+2) \, dx = x\ln(x+2) - \int x \frac{1}{x+2} \, dx\]
This method helps simplify the expression into a form where basic integration techniques can take over, leading toward solving the problem.

The natural logarithm, \( \ln(x) \), is a fundamental function in mathematics, commonly arising in calculus, especially when working with exponential growth and decay, and in integration issues. It is important to understand:- \( \ln(x) \) is defined only for \( x > 0 \).- The function is continuous and differentiable everywhere in its domain.

In the original exercise, our integral of interest is \( \ln(x+2) \). By shifting the function horizontally, we essentially move the entire graph of \( \ln(x) \) two units to the right. This transformation affects where the function is defined, starting from \( x > -2 \) rather than \( x > 0 \). Thus, for \( \ln(x+2) \), the input \( x+2 \) must be positive. In our application, the domain of interest is from \( x = -1 \) to \( x = 1 \), which is valid as \( x+2 \) will be positive across that entire range.
Understanding this function's domain and its behavior helps when calculating antiderivatives and using techniques like integration by parts.

The substitution method, much like a more intuitive change of variables, is a powerful tool that helps simplify the process of integration. It works especially well when dealing with composite functions. - We use substitution to transform the integral into a more manageable form.- Generally involves substituting \( w = g(x) \), and expressing \( dx \) in terms of \( dw \).

In the original integration, following the integration by parts application, the substitution \( w = x + 2 \) is used to solve the integral \( \int x \frac{1}{x+2} \, dx \). Substituting gives:\[w = x + 2 \implies dw = dx\]So, the integral becomes \( \int \frac{w-2}{w} \, dw \). Simplifying further:\[\int 1 - \frac{2}{w} \, dw = w - 2\ln|w| + C\]
Transforming back into terms of \( x \), we achieve a form that can be directly inserted into the integration by parts outcome. This substitution allows us to complete the evaluation of the integral effectively and elegantly.