Question

Evaluate the following integrals. Include absolute values only when needed. $$\int \frac{{\ln }^{2}x+2\ln x-1}{x}dx$$

Short answer

Question: Evaluate the integral: $$\int \frac{{\ln }^{2}x+2\ln x-1}{x}dx$$ Answer: $$\int \frac{{\ln }^{2}x+2\ln x-1}{x}dx=(\ln |x|({\ln }^{2}x+2\ln x-1)-2{\ln }^{2}x\ln |x|-2(\ln x{)}^2)+4\int \ln x\ln |x|dx+C$$

Step-by-step solution

Choose u and dv

Choose $u$ and $dv$ for integration by parts: $$u={\ln }^{2}x+2\ln x-1$$ $$dv=\frac{1}{x}dx$$

Find du and v

Find $du$ and $v$ by differentiating $u$ and integrating $dv$: $$du=(2\ln x+2)\frac{1}{x}dx$$ $$v=\int \frac{1}{x}dx=\ln |x|$$

Apply integration by parts formula

Apply the integration by parts formula, $\int udv=uv-\int vdu$: $$\int \frac{{\ln }^{2}x+2\ln x-1}{x}dx=({\ln }^{2}x+2\ln x-1)\ln |x|-\int (\ln |x|)(2\ln x+2)\frac{1}{x}dx$$

Simplify and distribute the integral

Simplify and distribute the integral: $$\int \frac{{\ln }^{2}x+2\ln x-1}{x}dx=\ln |x|({\ln }^{2}x+2\ln x-1)-2\int ({\ln }^{2}x)\frac{1}{x}dx-2\int (\ln x)\frac{1}{x}dx$$

Evaluate the remaining integrals

Evaluate the remaining integrals: For the first integral: $$-2\int ({\ln }^{2}x)\frac{1}{x}dx$$ By applying integration by parts again with: $$u={\ln }^{2}x$$ $$dv=\frac{-2}{x}dx$$ $$du=\frac{4\ln x}{x}dx$$ $$v=-2\ln |x|$$ We get: $$-2\int ({\ln }^{2}x)\frac{1}{x}dx=-2{\ln }^{2}x\ln |x|+4\int \ln x\ln |x|dx$$ For the second integral: $$-2\int (\ln x)\frac{1}{x}dx=-2(\ln x{)}^2+C$$

Combine the results and simplify

Combine the results and simplify the expression: $$\int \frac{{\ln }^{2}x+2\ln x-1}{x}dx=\ln |x|({\ln }^{2}x+2\ln x-1)-2{\ln }^{2}x\ln |x|+4\int \ln x\ln |x|dx-2(\ln x{)}^2+C$$ Since the term $4\int \ln x\ln |x|dx$ does not have a straightforward antiderivative, our final answer is: $$\int \frac{{\ln }^{2}x+2\ln x-1}{x}dx=(\ln |x|({\ln }^{2}x+2\ln x-1)-2{\ln }^{2}x\ln |x|-2(\ln x{)}^2)+4\int \ln x\ln |x|dx+C$$

Key concepts

Logarithmic Integration

Logarithmic integration involves integrating functions that include a logarithm such as $\ln x$.
This process can often be complex since the integration of logarithms isn't straightforward.

• Logarithmic integration often involves additional techniques like substitution or integration by parts.
• As seen in the solved exercise, we handle terms like ${\ln }^{2}x$ and $2\ln x$ while integrating.
• We ensure absolute values in logarithmic terms to handle negative inputs, though they're relevant only when the context demands.

Understanding the behavior of logarithms is crucial in accurately unfolding these integrals during mathematical evaluations.

Definite Integrals

Definite integrals are used to calculate the net area under a curve over a certain interval. While this particular problem focuses on indefinite integrals, it's useful to connect both concepts.

• Definite integrals have boundaries $[a,b]$ and yield a specific numerical value.
• These integrals play a key role in determining total quantities such as area, displacement, or accumulated quantities.
• The Fundamental Theorem of Calculus connects antiderivatives with definite integrals and helps evaluate them efficiently.

Although this exercise doesn't specifically require definite integrals, understanding them broadens comprehension when tackling similar integration problems.

Antiderivative

The antiderivative, also known as the indefinite integral, is a fundamental concept in calculus used to reverse the process of differentiation.

• Finding an antiderivative is essentially solving for a function whose derivative matches a given function.
• In the exercise, we found the antiderivative of $\frac{{\ln }^{2}x+2\ln x-1}{x}$, resulting in a complex expression.
• Antiderivatives add an arbitrary constant, $+C$, because differentiation eliminates constants in the original function.

Mastering antiderivatives is essential for evaluating both indefinite and definite integrals.

Integration Techniques

Integration techniques are strategies used to solve integrals beyond simple antiderivatives.

• Integration by Parts: This technique is based on the product rule for differentiation and is used extensively in the exercise.
• Substitution: Useful for transforming integrals into simpler forms, though not used in this particular instance.
• Trigonometric Integrals and Partial Fractions: Other advanced methods that handle specific types of integrands.

The exercise dominantly utilizes Integration by Parts. This involves the choices of functions $u$ and $dv$, transforming the original integral into a more accessible form. Utilization of these techniques reveals intricate relationships between function components and aids in determining complex integrals.