∫(2+x^2)/(1+x^2)dx
=∫(1+x^2+1)/(1+x^2)dx
=∫[1+1/(1+x^2)]dx
=x+arctanx+C
∫(1+2x)/(1+x^2)dx
=∫[1/(1+x^2)+2x/(1+x^2)]dx
=∫1/(1+x^2)dx+∫2x/(1+x^2)dx
=∫1/(1+x^2)dx+∫1/(1+x^2)d(1+x^2)
=arctanx+ln(1+x^2)+C
如果不懂,请追问,祝学习愉快!
追问谢谢了!
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/d8f9d72a6059252d694fb951379b033b5bb5b90f?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
那像这个呢?
追答你哪里的啊,怎么题目全英文的啊
∫(1+tanx)/secxdx
=∫(sinx+cosx)dx
=sinx-cosx+C
追问这么做对不对?
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/d6ca7bcb0a46f21f1035f806f5246b600c33ae66?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
追答对的。
追问谢谢!
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