Question

In the following exercises, verify by differentiation that ?lnxdx=x(lnx?1)+C, then use appropriate changes of variables to compute the integral. $$ \int \frac{\ln x}{\sqrt{x}} d x(H i n t: \operatorname{Set} u=\sqrt{x} .) $$

Short answer

The integral evaluates to \( 2\sqrt{x} \ln(x) - 4\sqrt{x} + C \).

Step-by-step solution

Interpret the Integral

We are given the integral \( \int \frac{\ln x}{\sqrt{x}} \, dx \) and have been advised to use a substitution where \( u = \sqrt{x} \). This suggests we should transform the variable from \( x \) to \( u \) to simplify the integral.

Make the Substitution

Let \( u = \sqrt{x} \), which implies \( x = u^2 \). Therefore, \( dx = 2u \, du \). Substitute these expressions into the integral to get:\[ \int \frac{\ln(u^2)}{u} \, (2u \, du)\]Simplifying this, the integral becomes \( \int 2 \ln(u^2) \, du \).

Simplify the Expression

Use the logarithmic identity \( \ln(u^2) = 2 \ln(u) \), so:\[\int 2 \ln(u^2) \, du = \int 4 \ln(u) \, du\]

Integrate with Respect to u

The integral \( \int 4 \ln(u) \, du \) can be solved using integration by parts. Set \( v = 4\ln(u) \) and \( dw = du \). Thus, \( dv = \frac{4}{u} \, du \) and \( w = u \). Applying integration by parts formula \( \int v \, dw = vw - \int w \, dv \), we have:\[4 \ln(u) \cdot u - \int u \cdot \frac{4}{u} \, du\]This becomes:\[4u \ln(u) - \int 4 \, du = 4u \ln(u) - 4u + C\]

Substitute Back to Original Variable

Replace \( u \) with \( \sqrt{x} \) to express the antiderivative in terms of \( x \):\[4\sqrt{x} \ln(\sqrt{x}) - 4\sqrt{x} + C\]Using \( \ln(\sqrt{x}) = \frac{1}{2}\ln(x) \), the final solution is expressed as:\[2\sqrt{x} \ln(x) - 4\sqrt{x} + C\]

Verify by Differentiation

Differentiate the expression \( 2\sqrt{x} \ln(x) - 4\sqrt{x} + C \) with respect to \( x \):\[\frac{d}{dx}[2\sqrt{x} \ln(x)] = \frac{2}{\sqrt{x}} \cdot \ln(x) + \sqrt{x}\cdot\frac{2}{x}\]\[\frac{d}{dx}[-4\sqrt{x}] = -\frac{2}{\sqrt{x}}\]Thus, the derivative is \( \frac{\ln(x)}{\sqrt{x}} \), confirming the integration is correct.

Key concepts

Integration by Parts

Integration by Parts is a technique used to solve integrals that involve products of functions. It is particularly useful when dealing with logarithmic or exponential functions combined with polynomials or other functions. The formula for integration by parts is derived from the product rule for differentiation and is given by:\[ \int u \, dv = uv - \int v \, du \]To apply this technique, we need to identify two parts of the integral: a function \( u \) and its differential \( dv \). Then, we differentiate \( u \) to get \( du \), and integrate \( dv \) to obtain \( v \).

• Choose \( u \) such that it becomes simpler when differentiated.
• Choose \( dv \) such that it can be easily integrated.

Different choices for \( u \) and \( dv \) may lead to different paths of simplification, but the result should remain consistent. In this problem, we set \( v = 4 \ln(u) \) and \( dw = du \) during the integration, simplifying the resulting expression so we could continue towards finding the antiderivative.

Logarithmic Functions

Logarithmic Functions are fundamental in calculus, often requiring special rules for integration and differentiation. The natural logarithm \( \ln(x) \) is the most common logarithmic function you'll encounter, especially when integrating functions involving the natural log directly or indirectly.One of the key properties of logarithmic functions used in integration is the identity:\[ \ln(a^b) = b \ln(a) \]This property allows us to simplify expressions like \( \ln(u^2) \) into \( 2 \ln(u) \), as seen in this exercise, streamlining the process of integration. Logarithms can be tricky because they change the rules of operation involving functions, so always keep the identities in mind.Also, be aware that integrating expressions involving \( \ln(x) \) often involves additional steps, such as integration by parts, due to their complex nature. Understanding these properties and strategies is crucial for effectively tackling calculus problems involving logarithmic functions.

Differentiation Verification

Differentiation Verification is a crucial step in calculus that ensures the integrity of your results after integration. When you solve an integral, particularly a complex one involving substitutions or parts, verifying by differentiation helps confirm that you've found the correct antiderivative.In the exercise, after integrating and obtaining the final form \( 2\sqrt{x} \ln(x) - 4\sqrt{x} + C \), we differentiate this expression with respect to \( x \). The process involves applying both the product and chain rules of differentiation because our solution includes products and compositions of functions. By finding that the derivative indeed simplifies back to \( \frac{\ln(x)}{\sqrt{x}} \), we ensure that the integration process was correctly executed. Differentiation acts as a kind of error-checking mechanism in calculus, reassuring us that the steps taken were mathematically sound.