Zanima me da li sam tačno našao koliko je [inlmath]\displaystyle\int\limits_{-1}^1\frac{\mathrm dx}{\left(x^2+1\right)\left(e^x+1\right)}[/inlmath]? Moje rešenje je sledeće:
[dispmath]\int\limits_{-1}^1\frac{\mathrm dx}{\left(x^2+1\right)\left(e^x+1\right)}=\left|u=\frac{1}{x^2+1},\;\mathrm du=\frac{-2x\,\mathrm dx}{\left(1+x^2\right)^2},\;\mathrm dv=\frac{\mathrm dx}{e^x+1},\;v=x-\ln\left(e^x+1\right)\right|=(*)\\
v=\int\frac{\mathrm dx}{e^x+1}=\left|t=e^x+1,\;\mathrm dt=e^x\,\mathrm dx\right|=\int\frac{\mathrm dt}{(t-1)t}=\int\left(\frac{A}{t-1}+\frac{B}{t}\right)\mathrm dt=(**)[/dispmath] Nakon sređivanja zadnjeg izraza u [inlmath](**)[/inlmath], dobio sam da je [inlmath]A=1[/inlmath], [inlmath]B=-1[/inlmath], pa kada sam to uvrstio u [inlmath](**)[/inlmath] dobio sam da je [inlmath]v=x-\ln\left(e^x+1\right)[/inlmath]. Sada za [inlmath](*)[/inlmath] važi:
[dispmath]\left.\frac{x-\ln\left(e^x+1\right)}{1+x^2}\right|_{-1}^1-\int\limits_{-1}^1\frac{\bigl(x-\ln\left(e^x+1\right)\bigr)(-2x)\,\mathrm dx}{\left(1+x^2\right)^2}=\frac{1}{2}+\int\limits_{-1}^1\frac{2x^2-2x\ln\left(e^x+1\right)}{\left(x^2+1\right)^2}\,\mathrm dx=\\
=\frac{1}{2}+\int\limits_{-1}^1\frac{x2x\,\mathrm dx}{\left(x^2+1\right)^2}-\int\limits_{-1}^1\frac{2x\ln\left(e^x+1\right)\,\mathrm dx}{\left(x^2+1\right)^2}=\left|t=x^2+1,\;\mathrm dt=2x\,\mathrm dx,\;t(-1)=2,\;t(1)=2\right|=\\
=\frac{1}{2}+\int\limits_2^2\frac{\sqrt{t-1}\,\mathrm dt}{t}-\int\limits_2^2\frac{\ln\left(e^{\sqrt{t-1}}+1\right)\mathrm dt}{t^2}=\frac{1}{2}+0-0=\frac{1}{2}.[/dispmath]
Mislim da sam tačno uradio, ali, nažalost, nisam siguran.