- Solution: \(\ln (x^2 - 1) = \ln (x + 1)(x - 1) = \ln (x + 1) + \ln (x - 1)\)
- \(\displaystyle \int x^n \ln(x) ~dx = \color{blue} { \frac {x^{n + 1}}{n + 1} (\ln (x) - \frac {1}{n + 1})}\)
Solution: \(u = \ln (x)\), \(v = \frac {x^{n + 1}}{n + 1}\), \(du = \frac {1}{x} ~dx\), \(dv = x^n ~dx\)
- \(\displaystyle \int_{1}^{\infty} \frac {\ln(x)}{x^2} ~dx = \color{blue} {1}\)
Solution: \(\frac {x^{(-2) + 1}}{(-2) + 1} (\ln (x) - \frac {1}{(-2) + 1}) = - \frac {\ln (x) + 1}{x}\)
Solution: \(\lim_{x \to \infty} \frac {\ln (x) + 1}{x} = \lim_{x \to \infty} \frac {\frac {1}{x} + 0}{1} = 0\) (L’Hospital Rule)
2. \(\displaystyle \int \sqrt {x} \ln(x) ~dx = \color{blue} {\frac {2}{3} x^{\frac {3}{2}} \ln (x) - \frac {4}{9} x^{\frac {3}{2}}}\)
- Solution: \(\frac {x^{\frac {1}{2} + 1}}{\frac {1}{2} + 1} (\ln (x) - \frac {1}{\frac {1}{2} + 1}) = \frac {2}{3} x^{\frac {3}{2}} (\ln (x) - \frac {2}{3}) \)
3. \(\displaystyle \int x^2\ln(x) ~dx = \color{blue} {\frac {1}{3} x^3 \ln (x) - \frac {1}{9} x^3}\)
- Solution: \(\frac {x^{2 + 1}}{2 + 1} (\ln (x) - \frac {1}{2 + 1}) = \frac {x^3}{3} (\ln (x) - \frac {1}{3})\)
4. \(\displaystyle \int x \ln(x) ~dx = \color{blue} {\frac {1}{2} x^2 \ln (x) - \frac {1}{4} x^2}\)
Solution: \(\frac {x^{1 + 1}}{1 + 1} (\ln (x) - \frac {1}{1 + 1}) = \frac {x^2}{2} (\ln (x) - \frac {1}{2})\)
Used 2018
5. \(\displaystyle \int \frac {\ln(x)}{x} ~dx = \color {blue} {\frac {1}{2} \ln ^2 (x)}\)
Solution: \(u = \ln (x)\), \(du = \frac {1}{x} ~dx\)
THIS IS DIFFERENT FROM THE REST (u-sub)
Solution
Take \(n \neq -1\) and \(x > 0\).
1. Integrate by parts with \(u = \ln x\) and \(dv = x^n\,dx\).
1.1. \(du = \frac{1}{x}\,dx\) and \(v = \frac{x^{n+1}}{n+1}\).
differentiate \(\ln x\) and integrate \(x^n\).
1.2. \(\int x^n \ln x\,dx = \frac{x^{n+1}}{n+1}\ln x - \int \frac{x^{n+1}}{n+1}\cdot \frac{1}{x}\,dx\).
integration by parts: \(\int u\,dv = uv - \int v\,du\).
1.3. \(= \frac{x^{n+1}}{n+1}\ln x - \frac{1}{n+1}\int x^n\,dx\).
\(\frac{x^{n+1}}{x} = x^n\).
1.4. \(= \frac{x^{n+1}}{n+1}\ln x - \frac{x^{n+1}}{(n+1)^2} + C\).
\(\int x^n\,dx = \frac{x^{n+1}}{n+1}\).
2. \(\int x^n \ln x\,dx = \frac{x^{n+1}}{n+1}\left(\ln x - \frac{1}{n+1}\right) + C\).
Factor out \(\frac{x^{n+1}}{n+1}\) in step 4 of step 1.
3. Q.E.D.
2 is the general formula; the special case \(n = -1\) is \(\int \frac{\ln x}{x}\,dx = \frac12 \ln^2 x + C\) (substitute \(u = \ln x\)).