Evaluate the integrals in Exercises 25–30 by using a substitution prior to integration by parts.
∫ ln(x + x²) dx

Verified step by step guidance

1. Start by examining the integral \(\int \ln(x + x^{2}) \, dx\). Notice that the argument of the logarithm can be factored to simplify the expression inside the logarithm.
2. Factor the expression inside the logarithm: \(x + x^{2} = x(1 + x)\). Use the logarithm property \(\ln(ab) = \ln a + \ln b\) to rewrite the integral as \(\int \ln x + \ln(1 + x) \, dx\).
3. Split the integral into two separate integrals: \(\int \ln x \, dx + \int \ln(1 + x) \, dx\). This makes the problem easier to handle by dealing with each term individually.
4. For each integral, use integration by parts. Recall the formula for integration by parts: \(\int u \, dv = uv - \int v \, du\). Choose \(u\) to be the logarithmic function (e.g., \(u = \ln x\) or \(u = \ln(1 + x)\)) and \(dv\) to be \(dx\).
5. Compute \(du\) and \(v\) for each integral: For \(u = \ln x\), \(du = \frac{1}{x} dx\); for \(dv = dx\), \(v = x\). Similarly, for \(u = \ln(1 + x)\), \(du = \frac{1}{1 + x} dx\); for \(dv = dx\), \(v = x\). Then apply the integration by parts formula to each integral separately.

Key Concepts

Substitution Method

The substitution method simplifies integrals by changing variables to transform the integral into a more manageable form. It involves choosing a substitution u = g(x) that simplifies the integrand, then rewriting the integral in terms of u and du. This technique is often used to prepare an integral for further methods like integration by parts.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation. It transforms the integral of a product of functions into simpler integrals using the formula ∫u dv = uv - ∫v du. Choosing u and dv wisely is crucial to simplify the integral effectively.

Logarithmic Integration

Integrals involving logarithmic functions often require special techniques, such as substitution or integration by parts, because logarithms do not have elementary antiderivatives in simple form. Recognizing how to manipulate the argument of the logarithm can simplify the integral and make it solvable.