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Evaluate $\int_{1}^{\infty} \frac{\ln (2 x - 1)}{x^{2}}$
My approach is to calc
$\int_{1}^{X} \frac{\ln (2 x - 1)}{x^{2}} d x$
and then take the limit for the answer when $X \to \infty$
However, I must do something wrong. The correct answer should be $2 \ln (2)$
$\int_{1}^{X} \frac{\ln (2 x - 1)}{x^{2}} d x = {[- \frac{1}{x} \ln (2 x - 1)]}_{1}^{X} + \int_{1}^{X} \frac{1}{x} \times \frac{2}{2 x - 1} d x = - \frac{1}{X} \ln (2 X - 1) + 2 \int_{1}^{X} - \frac{1}{x} + \frac{2}{2 x - 1} d x = - 1 \frac{1}{X} \ln (2 X - 1) - 2 \ln X + 2 \ln (2 X - 1)$
Am I wrong? If I'm not, how to proceed?
=== EDIT ===
After the edit I wonder if this is the correct way to proceed:
$- \frac{1}{X} \ln (2 X - 1) - 2 \ln X + 2 \ln (2 X - 1)$
The first part will do to zero because of $\frac{1}{X}$ so we ignore that one, the second and third part:
$- 2 \ln X + 2 \ln (2 X - 1) = 2 \ln (\frac{2 X - 1}{X}) = 2 \ln (2 - \frac{1}{X}) = 2 \ln (2)$

kennadiceKesezt 2022-09-19

Upper bound of natural logarithm
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Is Wolfram Alpha correct? If it is, why is it correct?
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Homework help to rearrange formula
Given the equation
$V_{m} = u (\ln m_{0} - \ln m_{8}) - g t_{f}$
I need to solve for $m_{0}$ Here is what I have but it looks messy and I feel like there is sometihng wrong or a better way
1st attempt
$\begin{aligned} V_{m} = u (\ln m_{0} - \ln m_{8}) - g t_{f} \\ V_{m} = u (\ln m_{0}) - u (\ln m_{8}) - g t_{f} \\ V_{m} = u (\ln m_{0}) - u (\ln m_{8}) - g t_{f} \\ V_{m} = u (\ln m_{0}) - u (\ln m_{8}) - g t_{f} \\ u (\ln m_{0}) - u (\ln m_{8}) - g t_{f} - V_{m} = 0 \\ u (\ln m_{0}) - u (\ln m_{8}) - V_{m} = g t_{f} \\ u (\ln m_{0}) = g t_{f} + u (\ln m_{8}) + V_{m} \\ \ln m_{0} = (g t_{f} + u (\ln m_{8}) + V_{m}) \div u \\ e^{(g t_{f} + u (\ln m_{8}) + V_{m}) \div u} = m_{0} \end{aligned}$
2nd attempt - think this looks a little better but still not there yet
$\begin{aligned} V_{m} = u (\ln \frac{m_{0}}{m_{8}}) - g t_{f} \\ V_{m} + g t_{f} = u (\ln \frac{m_{0}}{m_{8}}) \\ \frac{V_{m} + g t_{f}}{u} = \ln \frac{m_{0}}{m_{8}} \\ e^{\frac{V_{m} + g t_{f}}{u}} = \frac{m_{0}}{m_{8}} \\ m_{8} e^{\frac{V_{m} + g t_{f}}{u}} = m_{0} \end{aligned}$

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find f(-4)

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