Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int \frac{\ln x}{\sqrt{x}} d x $$

Short answer

\( \int \frac{\ln x}{\sqrt{x}} \, dx = 2\sqrt{x}(\ln x - 2) + C \)

Step-by-step solution

Identify Parts for Integration by Parts

To use integration by parts, we need to choose parts of the integral to differentiate and integrate. For \( \int \frac{\ln x}{\sqrt{x}} \, dx \), let \( u = \ln x \) (which will simplify when differentiated) and \( dv = \frac{1}{\sqrt{x}} \, dx \) (which is simpler to integrate).

Differentiate and Integrate Parts

Differentiate \( u \): \( du = \frac{1}{x} \, dx \). Integrate \( dv \): \( v = \int \frac{1}{\sqrt{x}} \, dx = 2\sqrt{x} \).

Apply Integration by Parts Formula

Use the integration by parts formula \( \int u \, dv = uv - \int v \, du \). Substitute in the expressions: \( \int \frac{\ln x}{\sqrt{x}} \, dx = \ln x(2\sqrt{x}) - \int 2\sqrt{x} \cdot \frac{1}{x} \, dx \).

Simplify and Evaluate Remaining Integral

Simplify the remaining integral: \( 2 \int \frac{1}{\sqrt{x}} \, dx = 2 \int x^{-1/2} \, dx \). Find this integral by increasing the exponent and dividing by it: \( 2 \cdot \left( \frac{x^{1/2}}{1/2} \right) = 2 \cdot 2\sqrt{x} = 4\sqrt{x} \).

Combine Results

The expression becomes \( 2\ln x \cdot \sqrt{x} - 4\sqrt{x} + C \). The complete solution to the integral is: \( \int \frac{\ln x}{\sqrt{x}} \, dx = 2\sqrt{x}(\ln x - 2) + C \).

Key concepts

Definite Integrals

Definite integrals help us find the area under a curve between two specified points. It contrasts with indefinite integrals that provide a general function family without specific boundaries. While our example deals with an indefinite integral, the process of integrating remains similar when dealing with definite integrals. The additional step in definite integrals involves evaluating the antiderivative at given points. This yields a precise value, often a measure of other physical quantities like distance or mass. When using integration by parts for a definite integral, remember to apply the evaluated limits to both the resulting terms of the integration by parts formula. Each evaluated term ensures precision. Understanding definite integrals is crucial for solving many real-life mathematical models that require exact calculations.

Differentiation Techniques

Differentiation techniques involve finding the derivative of a function. It is important in calculus because derivatives reveal rates of change and slopes of curves. In the original exercise, differentiation helps simplify the integral through the use of integration by parts.To use integration by parts, you need to correctly differentiate a portion of the integrand. In our example, the function \( u = \ln x \) is differentiated to find \( du = \frac{1}{x} \, dx \). This decision is strategic as it results in simplifying remaining components of the integral.Mastering differentiation techniques is invaluable because it aids in understanding how changes manifest in mathematical functions. This knowledge is fundamental before tackling integral calculus strategies like integration by parts.

Integrals Involving Logarithmic Functions

Integrals involving logarithmic functions can require advanced techniques like integration by parts to solve. Logarithmic components such as \( \ln x \) can often complicate direct integration. In the given exercise, choosing \( u = \ln x \) simplifies through differentiation, essential when using the integration by parts formula. Integration by parts helps manage the components, ensuring the intricate logarithmic function can be handled alongside other functions within the integral.When integrating logarithmic functions, consider their differentiability properties. The typical choice of setting \( u \) to a logarithmic part exploits these properties effectively, leading to simpler subsequent integrations.

Calculus Problem Solving

Calculus problem solving typically involves identifying techniques and tools that best address complex integral and differential questions. In our example, using integration by parts was crucial. Successfully solving calculus problems requires:

• Recognizing the type of integral or derivative needed.
• Choosing appropriate methods (such as integration by parts, substitution, or specific formulas).
• Understanding foundational calculus principles, including limits, continuity, and differentiability.
• Keeping track of assumptions and constraints, especially in applied contexts.

Successful problem solving often requires practice, as well as the ability to adapt known methods to novel challenges.