Question

Integration by parts to find the indefinite integral. $$ \int \ln x^{2} d x $$

Short answer

The integral of \(\int \ln x^{2} dx\) is hence \(2x \ln x - 2x + C\), where \(C\) is the constant of integration.

Step-by-step solution

Rewrite the Integral

Rewrite the integral \(\int \ln x^{2} dx\) as \(2 \int \ln x dx\). The rule \(\log a^b = b \log a\) has been used here to break the logarithmic function down into a simpler form.

Identify u and dv

For the method of integration by parts, the integral is broken up into two parts: 'u' and 'dv'. In this context, let 'u' be \(\ln x\), and 'dv' be \(dx\). The derivatives and antiderivatives become: 'du' = \(dx/ x\), and 'v' = \(x\).

Apply the Integration by Parts Formula

Substitute 'u', 'v', 'du', and 'dv' into the integration by parts formula \[ \int{u\, dv} = u.v - \int{v\, du} \]. This gives us \(2(x \ln x - \int x \cdot (dx / x))\). The integral \(\int x . dx / x\) simplifies to \(\int dx\), because \(x/x\) simplifies to 1.

Integral of dx

The integral \(\int dx\) is simply \(x\), which yields the final solution to the integral. Substituting this and simplifying the equation gives \(2(x \ln x - x)\). Which can further be written as \(2x \ln x - 2x\).

Key concepts

Indefinite Integral

Indefinite integrals might sound complex at first, but they are quite intuitive once you get familiar with them. They are used to find a family of functions where the derivative matches a given function. Imagine it as discovering the "anti-derivative" of a function.
The integral is called "indefinite" because it doesn't deal with specific limits or boundaries - instead, it gives a general form plus a constant, often represented as 'C'.

When you work with indefinite integrals, like \(\int \ln x^{2} dx\), you are essentially unraveling which function, when differentiated, would return the expression you start with. This process involves reverse engineering the differentiation rules you already know.

Integration Techniques

Integration by parts is a popular method used among various integration techniques to solve integrals that are products of functions. Its formula is derived from the product rule of differentiation, and it helps in breaking down complex integrations into simpler parts.
The formula is \[ \int{u\, dv} = u v - \int{v\, du} \]. The idea is to choose parts of the integrand (the function you're integrating) as 'u' (function we differentiate) and 'dv' (function we integrate).

In our problem \(\int \ln x dx\), we choose \(u = \ln x\) (since the derivative simplifies things here) and \(dv = dx\), making \(du = dx/x\) and \(v = x\). This transforms the integral into an easier one to solve. Picking the right 'u' and 'dv' can sometimes require trial and error, but you will become more confident as you practice.

Logarithmic Integration

Logarithmic integration often deals with functions involving logarithms, such as \(\ln x\). These integrals frequently require specific techniques like integration by parts because direct integration isn't straightforward.
For instance, integrating \(\ln x\) requires us to rewrite it in a way that fits the pattern of integration by parts. We identify parts that can simplify the logarithmic function into something integrable.

The step-by-step transformation we follow in your exercise not only simplifies the problem but also teaches critical problem-solving skills that are valuable in more complex calculus problems. Logarithms, by nature, turn multiplicative relationships into additive ones, making them crucial for breaking down exponential complexities during integration.