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}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v-x_1h9cgajAFfwHOP1q-gZO6gQxhkFw3yyCMynZcUeLLTTCAjMaLARQOR5m0QmqBlP5Z453jS1OuSOYXtfwy_8FGRDcBpgKqWDqieP_UjevQ_9IcKnQWSeqzgm75wSa1JtuTSwxbEDqWbcf7Id9M1BTMwUSq0L6w-kb1l_JoeopJkFd7f-pXFBQTHxPZzAVnCv3hfzlSilFHl_cfd87oWAsiefYxMSxow5cYK_t-_tbjKdT4ypMxmqNHLgbMPKFIzSJeXosiK8f3YJkgf1tMP-eoYRAFT1m5w8C9W-8AbDlKyBR3TjcEAzbXtRqUVpCxhKoNkN8s_xdV204b_fQLtZ5GPwST-ZA=s0-d "\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![\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}](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vyvjbiVv0oxPDOmp_D1ls5gJLC0e_Y5HGuXKTV4qYRMLsvDHwy2Pr2_vj06bS3SEvTj2GNgE1P6RtnMAdLhh3Vyaj1AuUISSmRkzybuDm7DlxOZOtcQ_z03uAPR5hGbhU4Dw4CImt9XJf9hP-BFwdXS6pOquMT_vYbxyBfWkJJhsNPtMqaKoOoFGyQD_H33FNP0JOa_B_dKG99qd5syGsjIaxwJv854pMMHfNLh7FbXVrafefsZuYRxmx612WSeCqmcW7Y8rYLkrCAC1D-IL4LI-ah3oyrPHIsRRV2T_FzMS_1qctT4l_YX69k5qPISPQA42jnOe6sDhj5K4EFEDrziGVl5eJ_4Jm-7_wc44n9yGRi4_-aUiUZo007l1OQmjqgMJu5VsGG4SBMxI5RV-gCKgG14RgKuOKZm6MVJ6v-oh_pbKc32XF2nt8SpFDyB-sTN3MeSc_QFtJMpyIt3Yaeu9Q8lTTmQBUr2ESWv-oRid1e9CXHxP3mIjHvg6z7BJ0GyKiNIBnItMPZnPtToeSq1Gp3tbOJov7ZBDmsnhrUuhLosrwpfqKAgLJJOEkUIdeZ_KK6bZtP-wstVUgjAQiWZtp_dHh3T_6mpTdkO5H5RyIYKuDcpiXVnWlTthA=s0-d "\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
