When to use Integration by Parts vs. U-Substitution

Deciding between integration by parts and u-substitution comes down to recognizing specific patterns in the integral.

For the integral $\int x \ln(1 + x) \,dx$, the correct method is integration by parts.

Here’s a breakdown of how to make that decision.

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Key Differences and When to Use Each Method

U-Substitution

The goal of u-substitution is to simplify the integral into a much more basic form, like $\int u^n \,du$ or $\int e^u \,du$.

Look for: A function and its derivative (or a constant multiple of its derivative) present in the integral.

• Classic Example: In $\int 2x \cos(x^2) \,dx$, you can see the function $x^2$ and its derivative $2x$. This is a perfect candidate for substitution where $u = x^2$.

In your problem, $\int x \ln(1 + x) \,dx$, the derivative of $\ln(1+x)$ is $\frac{1}{1+x}$, which isn't present. The derivative of $x$ is $1$, which doesn't help simplify the logarithm. Therefore, substitution is not the best first step.

Integration by Parts

This method is specifically designed to solve integrals that are a product of two unrelated functions. The formula is $\int u \,dv = uv - \int v \,du$.

Look for: A product of two different function types, such as:

• An algebraic function and a logarithmic function (like your problem)
• An algebraic function and an exponential function (e.g., $\int x e^x \,dx$)
• An algebraic function and a trigonometric function (e.g., $\int x \sin(x) \,dx$)

Your integral $\int x \ln(1 + x) \,dx$ is a textbook example of this pattern—it's the product of an algebraic function ($x$) and a logarithmic function ($\ln(1+x)$).

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How to Apply Integration by Parts to Your Problem

To choose which part is '$u$' and which is '$dv$', a helpful mnemonic is LIATE:

• Logarithmic
• Inverse Trigonometric
• Algebraic
• Trigonometric
• Exponential

Choose your '$u$' based on whichever function type appears first in the LIATE list.

1. Identify the function types:

   • $\ln(1+x)$ is Logarithmic.
   • $x$ is Algebraic.

2. Choose u: Since 'L' comes before 'A' in LIATE, we choose:

   • $u = \ln(1+x)$
   • $dv = x \,dx$

3. Differentiate u and integrate dv:

   • $du = \frac{1}{1+x} \,dx$
   • $v = \frac{x^2}{2}$

4. Apply the formula: $\int u \,dv = uv - \int v \,du$ $\int x \ln(1+x) \,dx = \frac{x^2}{2} \ln(1+x) - \int \frac{x^2}{2} \cdot \frac{1}{1+x} \,dx$ The new integral, $\int \frac{x^2}{2(1+x)} \,dx$, is now a rational function that can be solved with algebraic manipulation (like polynomial long division) and is simpler than the original problem.

Introducing terms

This technique involves introducing a "1" term into the integral to facilitate integration by parts.