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_tDyxNr7hmXvc-NTUAkNSsKwD7nsdcuGIKi-fFMh6qgfeuKMkD7_TKt7WW0cqyfFStNksq9W1D9HatZuZZfgRWciEOMD0osVX3suAzrYKCCSoAXU7ZSfuHE6h_uVvlDBcRhDlHppLA1k5I-COWaiysVz-nKLXLiEk9bRjNOAhh_v2tWEbOqkYUrqrp7WwoCs06ZBlqaL49Ea6TrPJYWVOc8oNtCM8Zehmgf6_lFq4cj3KgdP9_fhGvwBNImipx63bC1-uoM6pKWRUPp09KO0whSD-niCTpIFfo0NbfBMV5BCW-I0327S0jx_COku4m36GaxPD5hmDiUuZ_8nUxLeFLI2c1mi_4cfA=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_uy4IhRvRYACDxrxv_UQ1FQTXRIjXEnBazsf4SVeMqmdedXej3mU5jZxh2efbBCJYHKaRdoF0dNHeYm7IxXKZZgghsc-mZRbCZK7TMJrof8SDfz-azmwD-REWAbNUSlJwrKss6_BMkF54q7xYZ47ku-78VLLSADZTuu_mrBRttTfJXMJkfDQtdVQOiWRPix9A7rkIv61KHJLsMwpg0KU7PMz_6B6IWwFXpjKofY3_QI88cb-5tjuAQReSpMRObwSq8_GCgRDFzzPGCJtV1JisxQa8iwYNzk8cA3h2J49SVme8ID2Tu1c9ZGrv-Pzb3ggQ081Ccl3ruytpNor4Qez6i5r0f6JF7sJ79dVMJJknS9-hzPMFkoa9i6S-m5qG2v0E9P_bcwW5BXk2hLSd1Xetwo_0FQu0wfbcpawa5_jp9vkqf3QtAPMGwwVd4olT-dnGUZnkvN6Tmm2J-PeNXIt414Dh_uZ0U5Lj-l0jutHGYaMNPIXEpWK64OFbmGkuilWoEmcB3vidF9uw1pSIbVnxjGZaInTpjE6-xpmikTCLqATXgFmrOwICL1rhpp0bs6gBKQ7Ml1VwNkBvPRBxATiTxsMgFtWIYDOejqmJcPukYtvMJWqfdq0EHcNZbgU9c=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
