Integral s ln(x) na Grafičkom fakultetu
Na Grafičkom fakultetu u Zagrebu je na ispitu iz Matematike 2 bio zadan sljedeći zadatak:
Izračunajte integral
.Rješenje.
Izračunajmo prvo neodređeni integral; taj se lagano riješi metodom parcijalne integracije.
Stavimo li
![\begin{array}{lcl} u=\ln(1+x^2) & & dv=dx\\[15pt] \displaystyle du=\frac{2x}{1+x^2}dx & & v=x \end{array}](http://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Blcl%7D++u%3D%5Cln%281%2Bx%5E2%29+%26+%26+dv%3Ddx%5C%5C%5B15pt%5D++%5Cdisplaystyle+du%3D%5Cfrac%7B2x%7D%7B1%2Bx%5E2%7Ddx+%26+%26+v%3Dx++%5Cend%7Barray%7D&bg=ffffff&fg=333333&s=0 "\begin{array}{lcl} u=\ln(1+x^2) & & dv=dx\\[15pt] \displaystyle du=\frac{2x}{1+x^2}dx & & v=x \end{array}")
po formuli za parcijalnu integraciju 
imamo
imamo![\begin{array}{rcl} \displaystyle \int \ln(1+x^2) &=& \displaystyle x\ln(1+x^2) - 2\int\frac{x^2}{1+x^2}dx\\[15pt] &=& \displaystyle x\ln(1+x^2) - 2\int\frac{x^2+1-1}{1+x^2}dx\\[15pt] &=& \displaystyle x\ln(1+x^2) - 2x + 2{\rm arctg}\, x \end{array}](http://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Brcl%7D+%5Cdisplaystyle+%5Cint+%5Cln%281%2Bx%5E2%29++%26%3D%26+%5Cdisplaystyle+x%5Cln%281%2Bx%5E2%29+-+2%5Cint%5Cfrac%7Bx%5E2%7D%7B1%2Bx%5E2%7Ddx%5C%5C%5B15pt%5D++%26%3D%26+%5Cdisplaystyle+x%5Cln%281%2Bx%5E2%29+-+2%5Cint%5Cfrac%7Bx%5E2%2B1-1%7D%7B1%2Bx%5E2%7Ddx%5C%5C%5B15pt%5D++%26%3D%26+%5Cdisplaystyle+x%5Cln%281%2Bx%5E2%29+-+2x+%2B+2%7B%5Crm+arctg%7D%5C%2C+x++%5Cend%7Barray%7D&bg=ffffff&fg=333333&s=0 "\begin{array}{rcl} \displaystyle \int \ln(1+x^2) &=& \displaystyle x\ln(1+x^2) - 2\int\frac{x^2}{1+x^2}dx\\[15pt] &=& \displaystyle x\ln(1+x^2) - 2\int\frac{x^2+1-1}{1+x^2}dx\\[15pt] &=& \displaystyle x\ln(1+x^2) - 2x + 2{\rm arctg}\, x \end{array}")
pa je
