x = ln(t+1) => dx = \frac{dt}{t+1}| = \int\limits^4_{e-1} {\frac{dt}{t(t+1)} } = \int\limits^4_{e-1}(\frac{1}{t} - \frac{1}{t+1})dt = ln|t| - ln|t+1| = ln4 - ln5 - ln(e-1) + lne = ln0.8 - ln(e-1) + 1" alt="\int\limits^{2ln2}_1 {\frac{dx}{e^x-1}} = |e^x-1 = t => x = ln(t+1) => dx = \frac{dt}{t+1}| = \int\limits^4_{e-1} {\frac{dt}{t(t+1)} } = \int\limits^4_{e-1}(\frac{1}{t} - \frac{1}{t+1})dt = ln|t| - ln|t+1| = ln4 - ln5 - ln(e-1) + lne = ln0.8 - ln(e-1) + 1" align="absmiddle" class="latex-formula">