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**Soroban** Hello, totalnewbie!

Your substitution is off . . .

Substitute: .$\displaystyle \int\frac{\frac{2t}{1+t^2}\cdot\frac{2\,dt}{1+t^2} }{\frac{1-t^2}{1+t^2} - \frac{4t}{1+t^2} + 2}$ $\displaystyle = \;\int\frac{\frac{4t\,dt}{(1+t^2)^2}}{\frac{t^2 - 4t + 3}{1 + t^2}} $ $\displaystyle = \;\int\frac{4t\,dt}{(t-1)(t-3)(t^2+1)} $

Use partial fractions: .$\displaystyle \frac{4t}{(t-1)(t-3)(t^2+1)}\;=\;\frac{A}{t-1} + \frac{B}{t-3} + \frac{Ct + D}{t^2+1} $

. . I got: .$\displaystyle A = -1,\;B = \frac{3}{5},\;C = \frac{2}{5},\;D = -\frac{4}{5}$

Integrate: .$\displaystyle -\!\int\frac{dt}{t-1} \;+\;\frac{3}{5}\!\int\frac{dt}{t-3} \;+ \;\frac{2}{5}\!\int\frac{t\,dt}{t^2+1} \;- \;\frac{4}{5}\!\int\frac{dt}{t^2+1} $