求不定积分1/(2^x+5^x)dx 明天考试了，各位大佬帮帮忙啊，谢谢了 求不定积分∫1/[(x-1)^2(x+5)^2]dx详细过程
\u9ebb\u70e6\u5404\u4f4d\u5927\u4f6c\u5e2e\u5e2e\u5fd9\uff0c\u8c22\u8c22\u8c22\u8c22\u4e86\uff0c\u6c42\u4e0d\u5b9a\u79ef\u5206\u222b6x/(2\uff0b3x)dx\uff0c\u53e3\u7b97\u4e86\u4e00\u4e0b\uff0c\u6211\u6765\u7ed9\u9898\u4e3b\u4e00\u4e2a\u601d\u8def
\u5206\u5b506x\u6539\u4e3a(6x+4)-4\uff0c\u7136\u540e\u62c6\u5f00\u5c31\u53ef\u4ee5\u5f97\u51fa\u7b54\u6848\u4e86
\u5982\u679c\u8fd8\u4e0d\u4f1a\u7684\u8bdd\u4f60\u7559\u8a00\u4e0b\uff0c\u6211\u7ed9\u4f60\u5199\u51fa\u6765
=∫1/[(x+1)^2+4]dx
=∫(1/4)/[ [(x+1)/2]^2+1]dx
=∫(1/4)·2/[ [(x+1)/2]^2+1]d( (x+1)/2)
=(1/2)∫1/[ [(x+1)/2]^2+1]d( (x+1)/2)
=(1/2)arctan[(x+1)/2]+ C
拍题目图片
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https://zm12.sm-tc.cn/?src=l4uLj8XQ0JGKkIialoaK0YWeloeWnpHSmZ6RhpbRnJCS0JmekaCGlqDNx8nNz87O&uid=012c3bd86770e1cf3c085b91d26e4f3d&hid=895b155116d708b6b688605acaff9143&pos=1&cid=9&time=1547257029163&from=click&restype=1&pagetype=0000000000000402&bu=web&query=%E6%B1%82%EF%BC%8C%E5%AE%B6%E6%95%99%E7%8E%AFby%E9%92%A5%E5%8C%99%E7%8E%AF+%E5%86%99%E7%9A%84%E8%BF%99%E6%9C%AC%E6%96%87%EF%BC%8Cemmm%E8%83%BD%E4%B8%8D%E8%83%BD%E8%AF%B722%E7%AB%A0%E4%B9%9F%E7%BF%BB%E8%AF%91%E4%BA%86%E7%BB%99%E6%88%91&mode=&v=1&province=%E5%B1%B1%E4%B8%9C%E7%9C%81&city=%E6%B5%8E%E5%8D%97%E5%B8%82&uc_param_str=dnntnwvepffrgibijbprsvdsdichei
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