Problem 37 - College Algebra and Calculus: An Applied Approach

Question

Find the indefinite integral. (Hint: Integration by parts is not required for all the integrals.) $$ \int \frac{x e^{2 x}}{(2 x+1)^{2}} d x $$

Step-by-Step Solution

Answer

\(\frac{1}{2} [(2x + 1) e^{2x + 1} - e^{2x + 1}]\) + C, where C is the constant of integration.

Step 1: Identify the substitution target

Select \(u\) as \(2x + 1\). The differentiation of \(u\) which is \(du = 2dx\) will replace \(2x dx\) in the integral.

Step 2: Express the original integral in terms of u

Substitute \(u\) and \(du\) into the integral to get the new integral as \(\frac{1}{2} \int \frac{u e^{u}}{u^{2}} du\) which can be simplified to \(\frac{1}{2} \int \frac{e^{u}}{u} du\).

Step 3: Apply integration by parts

Select \(v = ln(u)\) and \(dw = e^{u} du\). Thus, \(dv = \frac{1}{u} du\) and \(w = e^{u}\). The integral becomes \(\frac{1}{2} [v \cdot w - \int w dv]\) which equals to \(\frac{1}{2} [u \cdot e^{u} - \int e^{u} \frac{1}{u} du]\).

Step 4: Solve the remaining integral

Solving \(\int e^{u} \frac{1}{u} du\), we have \(e^{u}\). Thus the previous step yields: \(\frac{1}{2} [u \cdot e^{u} - e^{u}]\).

Step 5: Back substitution

Substitute \(u\) by \(2x + 1\), our final solution will be \(\frac{1}{2} [(2x + 1) e^{2x + 1} - e^{2x + 1}]\). This is the solution to the indefinite integral.

Key Concepts

Substitution Method

The substitution method is a fundamental technique in integral calculus used to simplify integrals by changing the variable of integration. It is especially useful when an integral is too complex to solve directly, allowing us to rewrite it in terms of a simpler variable.
Let's break down this method:

• First, identify a part of the integral that can be substituted with a new variable, say \( u \). This usually involves selecting a portion of the integrand that, when differentiated, appears elsewhere in the integral.
• Next, differentiate \( u \) with respect to \( x \) to obtain \( du \). This step enables us to express \( dx \) in terms of \( du \).
• Rewrite the entire integral in terms of \( u \) and \( du \). This transformation often simplifies the integration process.
• Finally, integrate with respect to \( u \) and substitute back the original expression for \( u \). This gives the solution in the original variable.

In our exercise, we chose \( u = 2x + 1 \) because it appeared in the denominator, allowing us to express the integral as a function of \( u \) and solve it more easily.

Integration by Parts

Integration by Parts is another essential strategy in integral calculus, derived from the product rule of differentiation. It's particularly handy when the integrand is a product of two functions textually and the "u-substitution" technique cannot easily simplify it.
Consider the following steps:

• First, identify parts of the integrand to assign to \( v \) and \( dw \). Choose \( v \) such that its differentiation, \( dv \), and \( dw \) will not complicate the process.
• Then differentiate \( v \) to find \( dv \) and integrate \( dw \) to find \( w \).
• The formula for integration by parts is \( \int v \, dw = v \cdot w - \int w \, dv \). Begin by multiplying \( v \) and \( w \) and subtract the integral of \( w \) and \( dv \).

In our solution steps, we selected \( v = \ln(u) \) and \( dw = e^{u} du \), turning the challenging integral into an easier one involving just basic exponential and logarithmic functions.

Integral Calculus

Integral calculus is a fundamental part of calculus focused on the process of integration, aimed at finding the behaviors of functions over intervals. It involves finding the antiderivative or indefinite integral of a function, which essentially reverses differentiation, representing the accumulation of quantities.
Key aspects include:

• Indefinite integrals result in a function plus an arbitrary constant, since integration is an inverse operation of differentiation where the exact value of the original function at any point is unknown.
• The purpose of integration is to calculate areas, volumes, and other related quantities. By solving integrals, we understand the total accumulation of a particular function over a defined range.
• Diverse techniques like substitution and integration by parts are invaluable for solving complex integrals, simplifying them by breaking down components into easier parts.

Through these generalized methods, integral calculus allows us to tackle real-world problems regarding growth, area under curves, and total change, extending the range of practical applications.