Chapter 5: Problem 57
Use a computer algebra system to find or evaluate the integral. $$ \int \frac{\sqrt{x}}{x-1} d x $$
Short Answer
Expert verified
The integral of \(\frac{\sqrt{x}}{x-1} dx\) is \(\frac{2}{3} (x-1)^{1.5} + 2(x-1)^{0.5} + C\).
Step by step solution
01
Rewrite the integrand
Let's rewrite the given expression to simplify it, so the integrand becomes \(\frac{x^{0.5}}{x-1}\). This is a simple rewriting using the fact that square root of a number can be represented as that number raised to power 0.5.
02
Apply integral rule
Now it's time to take the integral. Note that \(x^{0.5}\) and \(x-1\) are both differentiable, which is a requirement for application of the integral rule - the rule of substitution. We need to choose a suitable function for substitution. Let's pick \(u = x - 1\), then \(du = dx\) and \(x = u + 1\). Therefore, our integral becomes: \(\int \frac{(u+1)^{0.5}}{u} du\).
03
Break Down the Expression and Evaluate the Integral
We can express the integral as the sum of two integrals as follows: \(\int \frac{(u+1)^{0.5}}{u} du = \int (u^{0.5} + u^{-0.5}) du = \int u^{0.5} du + \int u^{-0.5} du\). Use the power rule for integration, \(\int x^n dx = \frac{1}{n+1} \times x^{n+1} + C\) (with n not equal to -1) to solve. The constant of integration is denoted with C. Therefore, after we apply the power rule to both expressions, the evaluated integral is \(\frac{2}{3} u^{1.5} + 2u^{0.5} + C\).
04
Substitute Back the Value of u
Remember we made the substitution \(u = x - 1\). Let's go back to our original variable x. So, \(\frac{2}{3} u^{1.5} + 2u^{0.5} + C = \frac{2}{3} (x-1)^{1.5} + 2(x-1)^{0.5} + C\). So this is our final evaluated integral.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The Substitution Method is a powerful technique used in integral calculus to simplify integrals, especially when they appear too complex at first glance. It involves the following basic steps:
- Choosing a suitable substitution variable, often denoted by \( u \), to replace an expression in the integral.
- Finding the derivative of \( u \) with respect to \( x \), which helps in transforming the differential \( dx \) to \( du \).
- Substituting \( u \) and \( du \) into the original integral to obtain a simpler form.
- Evaluating the simpler integral and then replacing \( u \) back with the original expression in terms of \( x \).
Definite and Indefinite Integrals
Integrals in calculus come in two main types—definite and indefinite. Both types serve essential purposes in mathematics but differ in their functions and applications.
- Indefinite Integrals: These represent a family of functions and include an arbitrary constant \( C \), often written as \( \int f(x) \, dx = F(x) + C \). They are essential because they help find antiderivatives or reverse derivatives. The lack of limits means it captures general solutions.
- Definite Integrals: These have upper and lower limits of integration, such as \( \int_{a}^{b} f(x) \, dx \). This type provides specific numerical values representing the area under the curve from \( x=a \) to \( x=b \).
Power Rule for Integration
The Power Rule for Integration is a straightforward and vital tool in calculus for integrating functions of the form \( x^n \). It states that \[ \int x^n \, dx = \frac{1}{n+1} x^{n+1} + C, \]where \( n \) cannot be \( -1 \) since it would involve division by zero. This rule significantly simplifies the process of finding antiderivatives.
A few points to keep in mind when using the power rule:
A few points to keep in mind when using the power rule:
- Always ensure \( n eq -1 \) to avoid undefined expressions.
- Apply the rule separately to each term if the function is a polynomial.
- Remember to add the constant \( C \) for indefinite integrals to account for all possible antiderivatives.