- \(\displaystyle \int \ln (ax + b) ~dx = \color{blue} {\frac {ax + b}{a} \ln (ax + b) - x}\)
Solution: \(u = \ln (ax + b)\), \(v = x\), \(du = \frac {a}{ax + b} ~dx\), \(dv = dx\)
- \(\displaystyle \int \log _3 (x) ~dx = \color{blue} { \frac {x \ln (x) - x}{\ln (3)}}\)
Solution: \(\log _3 (x) = \frac {\ln (x)}{\ln (3)}\)
Used 2019
2. \(\displaystyle \int_{1}^{e} \ln (\sqrt {x}) ~dx = \color{blue} { \frac {1}{2}}\)
- Solution: \(\ln (\sqrt {x}) = \frac {1}{2} \ln (x)\)
3. \(\displaystyle \int \ln(x^2 + 6x + 5) ~dx = \color{blue} { (x + 1) \ln (x + 1) + (x + 5) \ln (x + 5) - 2x}\)
Solution: \(\ln(x^2 + 6x + 5)= \ln (x + 5)(x + 1) = \ln (x + 5) + \ln (x + 1)\)
Used 2018, Unsolved
Used 2019
4. \(\displaystyle \int \ln(x^2-1) ~dx = \color{blue} { (x + 1) \ln(x + 1) + (x - 1) \ln(x - 1) - 2x}\)
Solution
1. Evaluation of the base integral \(\int \ln(ax + b) \, dx\):
1.1. Substitute \(u = ax + b\), so \(du = a \, dx \implies dx = \frac{1}{a} \, du\).The integral becomes:\[ \int \ln(ax + b) \, dx = \frac{1}{a} \int \ln(u) \, du. \]
substitution method.
1.2. Integrating by parts with \(U = \ln u, \, dV = du \implies dU = \frac{1}{u} du, \, V = u\):\[ \int \ln u \, du = u \ln u - \int 1 \, du = u \ln u - u + C. \]
integration by parts.
1.3. Substituting back \(u = ax + b\):\[ \int \ln(ax + b) \, dx = \frac{ax + b}{a} \ln(ax + b) - \frac{ax + b}{a} + C = \frac{ax + b}{a} \ln(ax + b) - x + C. \]
absorbing the constant \(-\frac{b}{a}\) into \(C\).
2. Evaluation of sub-problems (1) through (4):
2.1. Sub-problem 1: \(\int \log_3(x) \, dx\)Using the change of base formula \(\log_3(x) = \frac{\ln x}{\ln 3}\):\[ \int \log_3(x) \, dx = \frac{1}{\ln 3} \int \ln x \, dx = \frac{x \ln x - x}{\ln 3} + C. \]
linearity of integration and 1 (<2>2).
2.2. Sub-problem 2: \(\int_1^e \ln(\sqrt{x}) \, dx\)Using the logarithm power rule \(\ln(\sqrt{x}) = \frac{1}{2} \ln x\):\[ \int_1^e \ln(\sqrt{x}) \, dx = \frac{1}{2} \int_1^e \ln x \, dx = \frac{1}{2} \Big[ x \ln x - x \Big]_1^e = \frac{1}{2} \Big( (e - e) - (0 - 1) \Big) = \frac{1}{2}(1) = \frac{1}{2}. \]
Fundamental Theorem of Calculus.
2.3. Sub-problem 3: \(\int \ln(x^2 + 6x + 5) \, dx\)Factoring \(x^2 + 6x + 5 = (x + 1)(x + 5)\):\[ \int \ln(x^2 + 6x + 5) \, dx = \int \ln(x + 1) \, dx + \int \ln(x + 5) \, dx. \]Applying 1 to each term gives:\[ \big((x + 1)\ln(x + 1) - x\big) + \big((x + 5)\ln(x + 5) - x\big) + C = (x + 1)\ln(x + 1) + (x + 5)\ln(x + 5) - 2x + C. \]
logarithm product rule and 1.
2.4. Sub-problem 4: \(\int \ln(x^2 - 1) \, dx\)Factoring \(x^2 - 1 = (x - 1)(x + 1)\):\[ \int \ln(x^2 - 1) \, dx = \int \ln(x - 1) \, dx + \int \ln(x + 1) \, dx = (x - 1)\ln(x - 1) + (x + 1)\ln(x + 1) - 2x + C. \]
logarithm product rule and 1.
3. Q.E.D.
Step 1 evaluates \(\int\ln(ax+b)\,dx\), and step 2 evaluates the four special cases.