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 in this site. In what follows, \( C \) is a constant of integration and can take any constant value.

Let \( U = \ln(x) \) and \( V' = 1 \) and use integration by parts. Hence \( U' = \dfrac{1}{x} \) and \( V = x \)

\[ \int \ln(x) \, dx = \int U V' \, dx \\\\ = U V - \int U' V \, dx \\\\ = x \ln(x) - \int 1 \, dx \\\ = x \ln(x) - x + C \] Check: Differentiate \( x \ln(x) - x + C \) and see that you obtain \( \ln(x) \) which is the integrand in the given integral. This is a way to check the answer to indefinite integrals evaluation.

Substitution: Let \( u = 2x + 1 \) which leads to \( \dfrac{du}{dx} = 2 \) or \( du = 2 \, dx \) or \( dx = \dfrac{du}{2} \), the above integral becomes \[ \int \ln(2x + 1) \, dx = \dfrac{1}{2} \int \ln(u) \, du \]

We now use integral formulas for \( \ln(x) \) (found in example 1) function to obtain \[ \int \ln(2x + 1) \, dx = \dfrac{1}{2} \int \ln(u) \, du = \dfrac{1}{2} [u \ln(u) - u] + C \] We now substitute \( u \) by \( 2x + 1 \) into the above to obtain \[ \int \ln(2x + 1) \, dx = \dfrac{1}{2}(2x + 1) \ln(2x + 1) - \dfrac{1}{2}(2x + 1) + C \] \[ = \dfrac{1}{2}(2x + 1) \ln(2x + 1) - x - \dfrac{1}{2} + C \] \[ = \dfrac{1}{2}(2x + 1) \ln(2x + 1) - x + K \] where \( K = C - \dfrac{1}{2} \) and is a constant.

Check: Differentiate \( \dfrac{1}{2}(2x + 1) \ln(2x + 1) - x + K \) and see that you obtain \( \ln(2x + 1) \) which is the integrand in the given integral. This is a way to check the answer to indefinite integrals evaluation.

Let \( f(x) = \ln x \) and \( g'(x) = x \) which gives \( f'(x) = \dfrac{1}{x} \) and \( g(x) = \dfrac{x^2}{2} \).

Using the integration by parts \[ \int f(x) g'(x) \, dx = f(x) g(x) - \int f'(x) g(x) \, dx \] , we obtain \[ \int x \ln x \, dx = \left[ \dfrac{x^2}{2} \ln x - \int \dfrac{x^2}{2} \cdot \dfrac{1}{x} \, dx \right] \] \[ = \dfrac{x^2}{2} \ln x - \int \dfrac{x}{2} \, dx \] \[ = \dfrac{x^2}{2} \ln x - \dfrac{x^2}{4} + C \].

Practice: Differentiate \( \dfrac{x^2}{2} \ln x - \dfrac{x^2}{4} + C \) to obtain the integrand \( x \ln x \) in the given integral.

Let \( u = \ln x \) so that \( \dfrac{du}{dx} = \dfrac{1}{x} \); after substitution, the given integral can be written as \[ \int \dfrac{\ln(x)}{x} \, dx = \int u \, du \] Integrate to obtain \[ \dfrac{u^2}{2} + C \] Substitute \( u \) by \( \ln x \) \[ = \left[\ln x\right]^2 / 2 + C \] As an exercise, check the final answer by differentiation.

1. \( \displaystyle \int x^3 \ln x \, dx \)

2. \( \displaystyle \int (x - \ln x) \, dx \)

2. \( -x \ln x + x^2 / 2 + x + C \)

- integrals and their applications in calculus.
- Integration by Parts