Problem 96 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{e^{x}}}\right) d x $$

Step-by-Step Solution

Answer

The indefinite integral is \( 2\sqrt{x} - 2e^{-x/2} + C \).

Step 1: Break Down the Integral

We begin by breaking down the integral into two separate integrals: \( \int \frac{1}{\sqrt{x}} \, dx + \int \frac{1}{\sqrt{e^x}} \, dx \). This allows us to compute each part individually.

Step 2: Integrate \( \frac{1}{\sqrt{x}} \)

Rewrite \( \frac{1}{\sqrt{x}} \) as \( x^{-1/2} \). The power rule for integration states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), for \( n eq -1 \). Apply this rule with \( n = -1/2 \): \[ \int x^{-1/2} \, dx = \frac{x^{1/2}}{1/2} + C_1 = 2\sqrt{x} + C_1 \].

Step 3: Simplify \( \frac{1}{\sqrt{e^x}} \)

Recognize that \( \frac{1}{\sqrt{e^x}} = e^{-x/2} \). This simplifies the integral.

Step 4: Integrate \( e^{-x/2} \)

For the exponential function \( e^{kx} \), the integral is \( \frac{1}{k} e^{kx} + C \). Here, \( k = -1/2 \), so: \[ \int e^{-x/2} \, dx = \frac{1}{-1/2} e^{-x/2} + C_2 = -2e^{-x/2} + C_2 \].

Step 5: Combine Results

Combine the results from Step 2 and Step 4: \( 2\sqrt{x} - 2e^{-x/2} + C \). Here, \( C = C_1 + C_2 \) is the constant of integration.

Key Concepts

Power Rule for Integration

Integrating functions can often seem daunting, but the power rule for integration simplifies the process for many polynomial expressions. This rule is essential when you encounter integrals of the form \( x^n \). The power rule states:

• For any real number \( n \) (except \( n = -1 \)), \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).
• The constant \( C \) represents the constant of integration, which accounts for all the possible vertical shifts of the antiderivative.

In our example with \( \int \frac{1}{\sqrt{x}} \, dx \), rewriting \( \frac{1}{\sqrt{x}} \) as \( x^{-1/2} \) allows us to apply the power rule easily. Substituting \( n = -1/2 \), the power rule yields:\[\int x^{-1/2} \, dx = \frac{x^{(1/2)}}{(1/2)} + C_1 = 2\sqrt{x} + C_1.\]This transformation is the backbone of solving integrals that can be expressed in polynomial form.

Exponential Function Integration

Exponential functions are common in calculus, and integrating them follows a straightforward pattern. If you have a function of the form \( e^{kx} \), the integration formula is:

• \( \int e^{kx} \, dx = \frac{1}{k} e^{kx} + C \)
• Here, \( k \) is a constant.

This formula results from the special property of the exponential function where derivatives and integrals have similar forms, with slight tweaks depending on the coefficient \( k \). In our problem, the expression \( \frac{1}{\sqrt{e^x}} \) simplifies to \( e^{-x/2} \). By setting \( k = -1/2 \), we apply the exponential function integration:\[\int e^{-x/2} \, dx = \frac{1}{-1/2} e^{-x/2} + C_2 = -2e^{-x/2} + C_2.\]This elegant pattern makes exponential integrals manageable, providing reliable results for complex calculations.

Integration of Radical Expressions

Radical expressions can often be daunting to integrate, but by using clever rewriting techniques, these expressions can turn into simpler forms. For radicals, which typically involve roots, rewriting the expression as a power of \( x \) or \( e \) significantly simplifies integration.

• For instance, \( \frac{1}{\sqrt{x}} \) is more approachable when rewritten as \( x^{-1/2} \).
• This allows us to utilize the power rule effectively.

Similarly, handling an expression like \( \frac{1}{\sqrt{e^x}} \) becomes simpler when recognized as \( e^{-x/2} \). This transformation leverages the properties of exponents and roots to apply known integration techniques.To handle such radical expressions:

• Always look to express the radical as a power.
• Determine if the function can be expressed in a form that fits known integral formulas like the power rule or exponential integration.

These strategies ease the complexity of integrating radical expressions, allowing for precise and clear solutions.