Question

Integrate: \(\int[(2 \mathrm{x}) /(\mathrm{x}+1)] \mathrm{dx}\).

Short answer

The integration of the given function \(\int[(2x)/(x+1)]dx\) is found by applying algebraic simplification and substitution methods. The final solution is \[2(x+1) - 2\ln|(x+1)|+C.\]

Step-by-step solution

Simplify the integrand

First, we can rewrite the given function in a slightly different form which will simplify the further calculations: \[\int\frac{2x}{x+1}dx = 2\int\frac{x}{x+1}dx\]

Apply the substitution method

Let's apply the substitution method. Let u = x + 1. Then, the derivative of u with respect to x is du/dx = 1, so dx = du. We'll rewrite our integral using this substitution: \[2\int\frac{x}{x+1}dx = 2\int\frac{u-1}{u}du\]

Simplify the integrand and integrate

Let's split up the fraction in the integrand: \[2\int\frac{u-1}{u}du = 2\int\left(\frac{u}{u} - \frac{1}{u}\right)du = 2\int(1 - u^{-1})du\] Now, we can integrate each term separately: \[2\int(1 - u^{-1})du = 2\int 1 du - 2\int u^{-1}du\] With the power rule of integration, we have: \[\int x^ndx = \frac{x^{(n+1)}}{n+1}+C\] Here n = -1 for the second term, so we will use the logarithm rule for integration: \[\int x^{-1}dx = \ln|x|+C\] Now, integrate the function: \[2\int 1 du - 2\int u^{-1}du = 2u - 2\ln|u|+ C\]

Substitute back

Now, remember that we let u = x + 1 at the beginning. We need to substitute back to express the answer in terms of x: \[2u - 2\ln|u|+ C = 2(x+1) - 2\ln|(x+1)|+C\] So, the final answer is: \[\int\frac{2x}{x+1}dx = 2(x+1) - 2\ln|(x+1)|+C\]

Key concepts

Substitution Method in Integration

The substitution method is a powerful tool in integration, especially useful when dealing with complex expressions. The idea is to simplify the integrand by substituting a part of it with a new variable. This technique is similar to finding the antiderivative of a composite function using the chain rule in reverse.

When applying the substitution method, typically, you'll choose a part of the integrand to be your 'u,' which is then differentiated to 'du.' This substitution simplifies the integral by transforming it into an easier form that often allows for straightforward integration. Let's take a step-by-step approach:

• Identify a portion of the integrand that, when differentiated, appears elsewhere in the integrand.
• Replace this portion with 'u,' and find 'du' by differentiating 'u' with respect to 'x.'
• Rewrite the integral in terms of 'u' and 'du,' simplifying where possible.
• Perform the integration on this simpler expression.
• Finally, substitute back the original variables to express the antiderivative in terms of the original variable.

For instance, in the exercise, we had \(\frac{2x}{x+1}dx\) which became much simpler when we applied the substitution \(u = x + 1\). This is a classic example of how substitution can turn a seemingly difficult integral into one that is much more manageable.

Integrating Rational Functions

Integrating rational functions, which are quotients of polynomials, can often be challenging. However, certain techniques, such as polynomial division or partial fraction decomposition, can make the task easier. In some cases, as seen in our exercise, the process can be significantly simplified by dividing the numerator by the denominator, if applicable.

In the given exercise, we observed that the integrand was a rational function. To integrate such functions, we sometimes look for a way to decompose the integrand into simpler parts. Here, the strategy included:

• Rewriting the integrand in a way that it could easily be separated into simpler fractions.
• Proceeding with the integration of each part, using rules and techniques appropriate for each term.

Breaking down the function into \(1 - u^{-1}\) allowed us to deal with a simple polynomial term and a term that led us to logarithmic integration – a perfect segue into the next concept.

Logarithmic Integration

Logarithmic integration directly addresses integrating functions of the form \(\frac{1}{x}\), which, unlike other power functions, do not follow the standard power rule for integration. Instead, they require a specific integration formula: \(\int\frac{1}{x}dx = \ln|x|+C\).

This special integration rule comes from the derivative of the natural logarithm function which is \(\frac{1}{x}\), and hence, the integral brings the logarithm back into the expression. In our exercise, after the substitution and breakdown of the integrand, we used the rule to integrate the \(u^{-1}\) term.

• This technique is particularly useful when the integral has a function that can be expressed as a derivative of a logarithmic function.
• Always remember the absolute value signs in the logarithmic function to ensure the solution is defined for all relevant x-values.

After integrating using the logarithmic rule, ensure that any substitutions made previously are reversed, returning the function to its original variable, as we did to provide the solution \(2(x+1) - 2\ln|(x+1)|+C\).