Evaluate integrals involving natural logarithmic functions: A tutorial, with examples and detailed solutions. Also exercises with answers are presented at the end of the tutorial. You may want to use the table of integrals and the properties of integrals on this site. In what follows, \( C \) is a constant of integration and can take any constant value.
Examples with Solutions
Click on each example to view its detailed solution.
Example 1: Evaluate \( \displaystyle \int \ln(x) \, dx \)
Solution to Example 1:
Let \( U = \ln(x) \) and \( V' = 1 \) and use integration by parts. Hence \( U' = \dfrac{1}{x} \) and \( V = x \):
\[ \begin{aligned} \int \ln(x) \, dx &= \int U V' \, dx \\ &= U V - \int U' V \, dx \\ &= x \ln(x) - \int 1 \, dx \\ &= x \ln(x) - x + C \end{aligned} \]Check: Differentiate \( x \ln(x) - x + C \) and see that you obtain \( \ln(x) \), which is the integrand in the given integral.
Example 2: Evaluate \( \displaystyle \int \ln(2x + 1) \, dx \)
Solution to Example 2:
Substitution: Let \( u = 2x + 1 \), which leads to \( \dfrac{du}{dx} = 2 \) or \( du = 2 \, dx \) or \( dx = \dfrac{du}{2} \). The integral becomes:
\[ \int \ln(2x + 1) \, dx = \dfrac{1}{2} \int \ln(u) \, du \]We now use integral formulas for the \( \ln(u) \) function (found in Example 1):
\[ \int \ln(2x + 1) \, dx = \dfrac{1}{2} \int \ln(u) \, du = \dfrac{1}{2} [u \ln(u) - u] + C \]Substitute \( u \) back by \( 2x + 1 \):
\[ \int \ln(2x + 1) \, dx = \dfrac{1}{2}(2x + 1) \ln(2x + 1) - \dfrac{1}{2}(2x + 1) + C \] \[ = \dfrac{1}{2}(2x + 1) \ln(2x + 1) - x - \dfrac{1}{2} + C \] \[ = \dfrac{1}{2}(2x + 1) \ln(2x + 1) - x + K \]where \( K = C - \dfrac{1}{2} \) is a constant.
Check: Differentiate the result to verify it matches the integrand.
Example 3: Evaluate \( \displaystyle \int x \ln x \, dx \)
Solution to Example 3:
Let \( f(x) = \ln x \) and \( g'(x) = x \), which gives \( f'(x) = \dfrac{1}{x} \) and \( g(x) = \dfrac{x^2}{2} \).
Using integration by parts \( \int f(x)g'(x) \, dx = f(x)g(x) - \int f'(x)g(x) \, dx \), we obtain:
\[ \begin{aligned} \int x \ln x \, dx &= \left[ \dfrac{x^2}{2} \ln x - \int \dfrac{x^2}{2} \cdot \dfrac{1}{x} \, dx \right] \\ &= \dfrac{x^2}{2} \ln x - \int \dfrac{x}{2} \, dx \\ &= \dfrac{x^2}{2} \ln x - \dfrac{x^2}{4} + C \end{aligned} \]Practice: Differentiate \( \dfrac{x^2}{2} \ln x - \dfrac{x^2}{4} + C \) to obtain the integrand \( x \ln x \).
Example 4: Evaluate \( \displaystyle \int \dfrac{\ln(x)}{x} \, dx \)
Solution to Example 4:
Let \( u = \ln x \) so that \( \dfrac{du}{dx} = \dfrac{1}{x} \) or \( du = \dfrac{1}{x} \, dx \). After substitution, the integral can be written as:
\[ \int \dfrac{\ln(x)}{x} \, dx = \int u \, du \]Integrate to obtain:
\[ \dfrac{u^2}{2} + C \]Substitute \( u \) back by \( \ln x \):
\[ = \dfrac{(\ln x)^2}{2} + C \]As an exercise, check the final answer by differentiation.
Exercises
Evaluate the following integrals:
- \( \displaystyle \int x^3 \ln x \, dx \)
- \( \displaystyle \int (x - \ln x) \, dx \)
Answers to Above Exercises
- \( \dfrac{x^4 \ln x}{4} - \dfrac{x^4}{16} + C \)
- \( \dfrac{x^2}{2} - x \ln x + x + C \)