Home » Blog » Find the integral from 0 to 1-e-2 of log(1-t) / (1-t)

Find the integral from 0 to 1-e-2 of log(1-t) / (1-t)

by RoRi

Evaluate the following integral:

    \[ \int_0^{1-e^{-2}} \frac{\log (1-t)}{1-t} \, dt. \]

We make the substitution u = \log (1-t), and du = \frac{-1}{1-t} \, dt. Under this substitution the new bounds of integration are

    \[ u(0) = \log (1) = 0, \qquad u(1-e^{-2}) = \log(1-(1-e^{-2})) = -2. \]

Therefore,

    \begin{align*} \int_0^{1-e^{-2}} \frac{\log(1-t)}{1-t} \, dt &= - \int_0^{-2} u \, du \\[10pt] &= -\frac{1}{2}u^2 \Big \rvert_0^{-2} \\[10pt] &= -2. \end{align*}

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):