Math Problem Statement

Evaluate the integral: ∫[1,9] 1/(√x(1 + √x)) dx

Solution

The given integral is:

191x(1+x)dx\int_1^9 \frac{1}{\sqrt{x}(1 + \sqrt{x})} \, dx

To solve this, we can make a substitution to simplify the expression.

Step 1: Substitution

Let u=xu = \sqrt{x}, which implies that x=u2x = u^2 and dx=2ududx = 2u \, du.

Now, when x=1x = 1, u=1u = 1, and when x=9x = 9, u=3u = 3.

Substitute x=u2x = u^2 into the integral:

191x(1+x)dx=131u(1+u)2udu\int_1^9 \frac{1}{\sqrt{x}(1 + \sqrt{x})} \, dx = \int_1^3 \frac{1}{u(1 + u)} \cdot 2u \, du

Step 2: Simplifying

The uu terms simplify:

21311+udu2 \int_1^3 \frac{1}{1 + u} \, du

Step 3: Integration

The integral of 11+u\frac{1}{1 + u} is ln(1+u)\ln(1 + u). Thus, we get:

2[ln(1+u)]132 \left[ \ln(1 + u) \right]_1^3

Step 4: Evaluate the Limits

Now, evaluate the definite integral:

[ 2 \left[ \ln(1 + 3) - \ln(1 + 1) \right] = 2 \left[ \ln(4) - \ln(2) \right] ]

Using the logarithmic identity ln(a)ln(b)=ln(ab)\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right), we get:

2ln(42)=2ln(2)2 \ln\left(\frac{4}{2}\right) = 2 \ln(2)

Finally, this simplifies to:

2ln(2)2 \ln(2)

Final Answer:

The value of the integral is:

2ln(2)2 \ln(2)

Would you like further clarification or details on any of the steps?

Here are 5 related questions to expand your understanding:

  1. How do logarithmic properties help simplify integrals?
  2. Can we apply a similar substitution for other forms of integrals?
  3. What is the geometric interpretation of the logarithmic function in this context?
  4. How would this integral change if the limits of integration were different?
  5. What are the conditions under which substitution is a helpful method for solving integrals?

Tip: Always check if a substitution will simplify both the differential and the integrand before proceeding with it.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Integration
Substitution Method
Logarithms

Formulas

Substitution: u = √x, x = u^2, dx = 2u du
Integration of 1/(1 + u) = ln(1 + u)
Logarithmic identity: ln(a) - ln(b) = ln(a/b)

Theorems

Fundamental Theorem of Calculus
Logarithmic Properties

Suitable Grade Level

Undergraduate Calculus