Problem 14 - Essential Calculus

Question

Find the integral. (Note: Solve by the simplest method-not all require integration by parts.) $$ \int \frac{\ln x}{x^{2}} d x $$

Step-by-Step Solution

Answer

So, the integral \(\int{\frac{\ln{x}}{x^{2}} dx}\) equals \(-\frac{\ln{x}}{x} - \frac{1}{x} + C\)

Step 1: Choosing 'u' and 'dv'

First set \(u = \ln{x}\), because the derivative of \(\ln{x}\) is simpler (\(\frac{1}{x}\)), and set \(dv = \frac{1}{x^{2}} dx\) because you can easily integrate it.

Step 2: Compute 'du' and 'v'

By differentiating 'u', you get \(du = \frac{1}{x} dx\), and integrating 'dv', you get \(v = -\frac{1}{x}\).

Step 3: Apply the Integration by Parts formula

Use the integration by parts formula \(\int{u dv} = u.v - \int{v du}\), which gives us: \(\int{\frac{\ln{x}}{x^{2}} dx} = \ln{x} \cdot -\frac{1}{x} - \int{-\frac{1}{x}}\cdot \frac{1}{x} dx\)

Step 4: Solve the Integral

Solve the remaining integral, \(\int{-\frac{1}{x}}\cdot \frac{1}{x} dx = \int{-\frac{1}{x^{2}}} dx = \frac{1}{x}\), and substitute this into the previous expression.

Step 5: Combine Results

Adding all the derived results provides the solution: \(\int{\frac{\ln{x}}{x^{2}} dx} = -\frac{\ln{x}}{x} - \frac{1}{x} + C\), where \(C\) is the constant of integration.