Problem 37 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int \frac{x}{\sqrt{1+x}} d x$$

Step-by-Step Solution

Answer

\( \int \frac{x}{\sqrt{1+x}} \, dx = \frac{2}{3} (1 + x)^{3/2} - 2 (1 + x)^{1/2} + C. \)

Step 1: Choose a Substitution

Let's simplify the integral by using a substitution. We'll set \( u = 1 + x \). Then, \( du = dx \) and \( x = u - 1 \). This substitution will make the integral easier to handle.

Step 2: Substitute into Integral

Substitute \( u = 1 + x \) and \( x = u - 1 \) into the integral. This gives us the new integral in terms of \( u \):\[ \int \frac{u-1}{\sqrt{u}} \, du. \]

Step 3: Simplify the Integral

Separate the integral into two simpler integrals: \[ \int \frac{u}{\sqrt{u}} \, du - \int \frac{1}{\sqrt{u}} \, du. \] Simplify each term: \( \frac{u}{\sqrt{u}} = \sqrt{u} \) and \( \frac{1}{\sqrt{u}} = u^{-1/2} \).

Step 4: Integrate Each Term

Now integrate each simplified term separately. For \( \int \sqrt{u} \, du \): Use the power rule for integration, \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). Here, \( n = 1/2 \), so \[ \int \sqrt{u} \, du = \frac{2}{3} u^{3/2} + C. \]For \( \int u^{-1/2} \, du \): Again using the power rule, where \( n = -1/2 \), so \[ \int u^{-1/2} \, du = 2 u^{1/2} + C. \]

Step 5: Combine and Back-Substitute

Combine the results from Step 4: \[ \frac{2}{3} u^{3/2} - 2 u^{1/2} + C. \]Now substitute back \( u = 1 + x \) to express the result in terms of \( x \):\[ \frac{2}{3} (1 + x)^{3/2} - 2 (1 + x)^{1/2} + C. \]

Step 6: Write Final Answer

The evaluated integral is: \[ \int \frac{x}{\sqrt{1+x}} \, dx = \frac{2}{3} (1 + x)^{3/2} - 2 (1 + x)^{1/2} + C. \]

Key Concepts

Integration

Integration is a fundamental concept in calculus that is used to find the accumulated quantity, such as areas under curves or the total change of a function. Unlike differentiation which deals with the rate of change, integration helps us combine pieces together, like summing infinitesimal parts to get a whole.
In the given problem, we are tasked with integrating the function \( \int \frac{x}{\sqrt{1+x}} \, dx \). This integral is tricky to solve using straightforward methods, so we employ more advanced techniques like substitution. This technique converts the given integral into a simpler form, making it easier to evaluate.
When performing integration, it's essential to remember the importance of constants. The constant \( C \) is added to the final result to represent any constant of integration that might have been lost during differentiation, as integration is an inverse operation to differentiation.

Substitution Method

The substitution method is a powerful tool in calculus used to simplify integrals by transforming the integrand into a form that is more easily integrable. The idea is to "substitute" part of the integral with a new variable \( u \). This is particularly useful when dealing with composite functions and is similar to the technique of change of variables.
In our example, we substituted \( u = 1 + x \), which transforms the function \( \int \frac{x}{\sqrt{1+x}} \, dx \) into a more tractable form \( \int \frac{u-1}{\sqrt{u}} \, du \). We then replace every occurrence of \( x \) in terms of \( u \), including changing \( dx \) to \( du \) based on our substitution. This simplifies the integration process by removing complexities associated with the original variables.
Choosing the right substitution can greatly influence the ease of solving an integral. Ideally, the new expression should unfold into simpler parts that can be integrated using basic rules or methods.

Power Rule for Integration

The power rule for integration is one of the simplest and most frequently used rules in calculus. It allows us to integrate functions of the form \( u^n \). According to the power rule, the integral of \( u^n \) with respect to \( u \) is given by \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), for all \( n eq -1 \).
In our problem, after using the substitution method, we apply the power rule to deal with the expressions \( \sqrt{u} \) and \( u^{-1/2} \). The power rule simplifies these terms to \( \int u^{1/2} \, du = \frac{2}{3} u^{3/2} + C \) and \( \int u^{-1/2} \, du = 2 u^{1/2} + C \) respectively.
The application of the power rule enables us to effortlessly solve parts of our transformed integral. This rule is fundamental because many complex integrals can be reduced to ones that apply the power rule directly or with slight modifications.