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模型定价
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在线使用
先点侧边栏的令牌，再点令牌的聊天，本站提供多个前端可选择
不要用playground，因为playground未适配部分模型的一些参数，容易报错
不要用侧边栏的，因为网络路由不稳定，请始终点击令牌后点击令牌的聊天，进入新网站 https://chat.bailili.top/#/ 后对话
可以选择不同的前端：
Next-Web-Langchain：默认前端，点击即可使用，无需额外配置，支持插件调用。相比于原版nextchat，功能更多，无广告
Lobelite文件端：支持上传PDF文件
Next-Web-Langchain
可选操作
可以适量调整上下文长度，和是否开启压缩历史消息（压缩时也会产生请求）
可以回到最上方选择是否自动生成标题（自动生成标题也会产生请求）
这些请求都可以在api网站的日志看到
改完所有设置后会自动保存，直接右上角叉掉就可以开始对话
插件模式画图
点击设置，选择dall-e-3，叉掉
选择模型，选择gpt-4o，打开插件模式
开启dalle插件
Lobe文件端
Lobe文件端：支持文件解析，文件存储，进入后先左上角使用github登录，然后点左上角应用设置-AI服务商-OpenaAI，API Key自己复制，API 代理地址填写 https://api.bailili.top/v1
获取模型列表，开启所需模型
语言模型填入key，获取模型列表，按需填入模型后即可对话（会使用嵌入模型，产生极小额外消费）
mj实时语音：进去后手动设置 url 为 https://new.bailili.top ，key自己复制，然后点语音
故障排除
如果报错，请检查排除下面的情况：
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模型介绍
gemini
gemini-2.5-pro-exp-03-25 训练知识截止：25年1月 发布时间：25年3月
最先进的思考模型，能够推理代码、数学和 STEM 领域的复杂问题，以及使用长上下文分析大型数据集、代码库和文档
问：最近更新了2.5Pro之后，输出截断更多了，似乎是内容审查导致的？怎么解决？
答：用newapi进行中转后，已设置为最低审查。需要注意的是，就算关闭了所有可以关闭的审查，csam等“基础审核”仍然存在，而且不可关闭，因此截断内容/空回复的现象仍然会存在，使用非流请求对输出截断有一定的缓解。
网友：可以要求他在尾部输出免责声明什么的，最后拿正则去了，Gemini的截断应该是外置模型审核，很好骗的
gemini-2.0-flash-thinking-exp 先前的实验性思考模型
gemini-2.0-flash 新一代功能、速度、思考、实时串流和多模式生成
gemini-2.0-flash-lite-preview 成本效益高且延迟时间短
gemini-1.5-pro-002 需要更高智能的复杂推理任务
更多参数参考Gemini 模型 | Gemini API | Google AI for Developers 注意，gemini所有的思考模型api都不输出过程
GCP速率限制 API速率限制 普号gemini-2.5-pro-exp-03-25：25RPD
在线使用速率限制 普号gemini-2.5-pro-exp-03-25：50RPD
查看API可用模型curl -X GET 'https://generativelanguage.googleapis.com/v1beta/models?key='
API调用
base url
请根据程序的参考示例尝试下面的地址：
本站api所有模型均为标准openai接口，需自行在令牌页生成key，然后将接口替换成本站：
https://api.openai.com -> https://api.bailili.top
https://api.openai.com/ -> https://api.bailili.top/
https://api.openai.com/v1 -> https://api.bailili.top/v1
https://api.openai.com/v1/chat/completions -> https://api.bailili.top/v1/chat/completions
请注意，很多网站要填的是第一行，无需加域名结尾的/
key
key就是令牌，请始终生成auto分组的key以访问所有模型
请 妥善保管 你的 key，不要将key填在不明的网站和应用内，以防被他人刷取费用！如有需要，可给令牌设置额度防止大额损失
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当令牌设为无限额度时，仅账户余额限制令牌使用额度
站点选择
使用域名时，如果有超过100的日志，前端却没有收到或显示524，是因为cf连接超过100就关闭了
o1-mini，claude3-7等模型在长文本下思考时间较长，可能经常超过100s，若有上述场景，建议使用流式请求，或联系客服使用ip
api接口基础教程
聊天请求：聊天接口（OpenAI格式通用） - AI模型API中文文档
嵌入请求：创建嵌入 - API reference
本站支持openai接口调用所有模型，支持claude原生格式调用claude，暂不支持gemini原生格式调用gemini
假如终端应用给出openai接口和gemini接口，应选择openai接口接入本站gemini模型
接入newapi二次分发
Calcium-Ion/new-api-horizon: 特性版 - 基于New API开发的增强版本，专注于高性能
本站使用horizon版本，也推荐下游使用
部署后，添加渠道，代理地址填写 https://api.bailili.top，密钥自己复制，即可接入
接入claude-3-7-sonnet-20250219
1.claude官方格式接入thinking
官方文档：Building with extended thinking - Anthropic
POST:https://api.bailili.top/v1/messages
{
"model": "claude-3-7-sonnet-20250219",
"max_tokens": 20000,
"thinking": {
"type": "enabled",
"budget_tokens": 16000
},
"messages": [
{
"role": "user",
"content": "1/(1+x^3)的不定积分"
}
]
}claude原生格式接入claude-3-7-sonnet-20250219不再赘述，上面的格式删掉"thinking"即可
{
"id": "msg_01LkSHA4Y6mWswxpyxRn7z7x",
"type": "message",
"role": "assistant",
"model": "claude-3-7-sonnet-20250219",
"content": [
{
"type": "thinking",
"thinking": "I need to find the indefinite integral of $\\frac{1}{1+x^3}$.\n\nThis is a non-trivial integral. Let's try to do a partial fraction decomposition. \n\nWe have $1 + x^3 = 0$ at $x = -1$.\n\nActually, let's factor $1 + x^3$. We have \n\\begin{align}\n1 + x^3 &= 1 + x^3\\\\\n\\end{align}\n\nLet's think more carefully. We know that if $x^3 = -1$, then $x = -1$ is one root. The other roots are complex, and we can find them by factoring.\n\n\\begin{align}\n1 + x^3 &= (x + 1)(x^2 - x + 1)\n\\end{align}\n\nWe can verify this: $(x+1)(x^2 - x + 1) = x^3 - x^2 + x + x^2 - x + 1 = x^3 + 1$.\n\nNow, we can do partial fraction decomposition.\n\\begin{align}\n\\frac{1}{1+x^3} &= \\frac{1}{(x+1)(x^2 - x + 1)}\\\\\n&= \\frac{A}{x+1} + \\frac{Bx + C}{x^2 - x + 1}\n\\end{align}\n\nMultiplying both sides by $(x+1)(x^2 - x + 1)$, we get \n\\begin{align}\n1 &= A(x^2 - x + 1) + (Bx + C)(x+1)\\\\\n&= A(x^2 - x + 1) + Bx(x+1) + C(x+1)\\\\\n&= A(x^2 - x + 1) + Bx^2 + Bx + Cx + C\\\\\n&= Ax^2 - Ax + A + Bx^2 + Bx + Cx + C\\\\\n&= (A+B)x^2 + (-A+B+C)x + (A+C)\n\\end{align}\n\nComparing with the coefficients of the left-hand side (which is just 1), we get \n\\begin{align}\nA + B &= 0\\\\\n-A + B + C &= 0\\\\\nA + C &= 1\n\\end{align}\n\nFrom the first equation, we have $B = -A$. Substitute into the second equation.\n\\begin{align}\n-A - A + C &= 0\\\\\n\\Rightarrow -2A + C &= 0\\\\\n\\Rightarrow C &= 2A\n\\end{align}\n\nSubstitute into the third equation.\n\\begin{align}\nA + 2A &= 1\\\\\n\\Rightarrow 3A &= 1\\\\\n\\Rightarrow A &= \\frac{1}{3}\n\\end{align}\n\nNow we can find the other variables.\n\\begin{align}\nB &= -A = -\\frac{1}{3}\\\\\nC &= 2A = \\frac{2}{3}\n\\end{align}\n\nSo, we have \n\\begin{align}\n\\frac{1}{1+x^3} &= \\frac{1/3}{x+1} + \\frac{-x/3 + 2/3}{x^2 - x + 1}\\\\\n&= \\frac{1}{3(x+1)} + \\frac{-x + 2}{3(x^2 - x + 1)}\n\\end{align}\n\nNow, we need to compute the integrals of each term.\n\nFor the first term, $\\frac{1}{3(x+1)}$, the integral is $\\frac{1}{3}\\ln|x+1| + C_1$.\n\nFor the second term, $\\frac{-x + 2}{3(x^2 - x + 1)}$, let's simplify.\n\\begin{align}\n\\frac{-x + 2}{3(x^2 - x + 1)} &= \\frac{-x + 1/2 + 3/2}{3(x^2 - x + 1)}\\\\\n&= \\frac{-(x - 1/2)}{3(x^2 - x + 1)} + \\frac{3/2}{3(x^2 - x + 1)}\\\\\n&= \\frac{-(x - 1/2)}{3(x^2 - x + 1)} + \\frac{1/2}{(x^2 - x + 1)}\n\\end{align}\n\nNow, let's consider each new term separately.\n\nFor $\\frac{-(x - 1/2)}{3(x^2 - x + 1)}$, we note that the derivative of $x^2 - x + 1$ is $2x - 1 = 2(x - 1/2)$. So, \n\\begin{align}\n\\frac{-(x - 1/2)}{3(x^2 - x + 1)} &= \\frac{-1}{6} \\cdot \\frac{2(x - 1/2)}{(x^2 - x + 1)}\\\\\n&= \\frac{-1}{6} \\cdot \\frac{d}{dx}[\\ln(x^2 - x + 1)]\n\\end{align}\n\nThe integral of this is $\\frac{-1}{6}\\ln(x^2 - x + 1) + C_2$.\n\nActually I'm making an error above because I'm getting confused with signs. Let's redo the calculation in a cleaner way.\n\nLet's reconsider the partial fraction decomposition.\nWe have:\n\\begin{align}\n\\frac{1}{1+x^3} &= \\frac{1}{(x+1)(x^2 - x + 1)}\\\\\n&= \\frac{A}{x+1} + \\frac{Bx + C}{x^2 - x + 1}\n\\end{align}\n\nMultiplying both sides by $(x+1)(x^2 - x + 1)$, we get \n\\begin{align}\n1 &= A(x^2 - x + 1) + (Bx + C)(x+1)\\\\\n\\end{align}\n\nWhen $x = -1$, we get:\n\\begin{align}\n1 &= A(1 + 1 + 1)\\\\\n&= 3A\\\\\n\\Rightarrow A &= \\frac{1}{3}\n\\end{align}\n\nTo find $B$ and $C$, let's expand the equation:\n\\begin{align}\n1 &= A(x^2 - x + 1) + (Bx + C)(x+1)\\\\\n&= \\frac{1}{3}(x^2 - x + 1) + (Bx + C)(x+1)\\\\\n&= \\frac{1}{3}x^2 - \\frac{1}{3}x + \\frac{1}{3} + Bx^2 + Bx + Cx + C\\\\\n&= (\\frac{1}{3} + B)x^2 + (-\\frac{1}{3} + B + C)x + (\\frac{1}{3} + C)\n\\end{align}\n\nComparing coefficients with the left-hand side:\n\\begin{align}\n\\frac{1}{3} + B &= 0\\\\\n-\\frac{1}{3} + B + C &= 0\\\\\n\\frac{1}{3} + C &= 1\n\\end{align}\n\nFrom the first equation, $B = -\\frac{1}{3}$.\nFrom the third equation, $C = \\frac{2}{3}$.\n\nLet's check the second equation:\n\\begin{align}\n-\\frac{1}{3} + (-\\frac{1}{3}) + \\frac{2}{3} &= 0\\\\\n-\\frac{1}{3} - \\frac{1}{3} + \\frac{2}{3} &= 0\\\\\n\\frac{-2 + 2}{3} &= 0\\\\\n0 &= 0\n\\end{align}\n\nGreat, so it checks out.\n\nSo, we have:\n\\begin{align}\n\\frac{1}{1+x^3} &= \\frac{1/3}{x+1} + \\frac{-x/3 + 2/3}{x^2 - x + 1}\\\\\n&= \\frac{1}{3(x+1)} + \\frac{-x/3 + 2/3}{x^2 - x + 1}\n\\end{align}\n\nNow, we need to integrate each term.\n\nFor the first term, $\\frac{1}{3(x+1)}$, the integral is $\\frac{1}{3}\\ln|x+1| + C_1$.\n\nFor the second term, $\\frac{-x/3 + 2/3}{x^2 - x + 1}$, let's split it:\n\\begin{align}\n\\frac{-x/3 + 2/3}{x^2 - x + 1} &= \\frac{-x/3}{x^2 - x + 1} + \\frac{2/3}{x^2 - x + 1}\n\\end{align}\n\nFor $\\frac{-x/3}{x^2 - x + 1}$, we can rewrite:\n\\begin{align}\n\\frac{-x/3}{x^2 - x + 1} &= \\frac{-1}{3} \\cdot \\frac{x}{x^2 - x + 1}\n\\end{align}\n\nThe derivative of $x^2 - x + 1$ is $2x - 1$. We can't directly use substitution.\n\nLet me rewrite the term from the beginning:\n\\begin{align}\n\\frac{-x/3 + 2/3}{x^2 - x + 1} &= \\frac{-x/3 + 1/6 + 1/2}{x^2 - x + 1}\\\\\n&= \\frac{-x/3 + 1/6}{x^2 - x + 1} + \\frac{1/2}{x^2 - x + 1}\n\\end{align}\n\nNotice that $-x/3 + 1/6 = -(x/3 - 1/6) = -\\frac{1}{3}(x - 1/2)$. And also notice that $\\frac{d}{dx}(x^2 - x + 1) = 2x - 1 = 2(x - 1/2)$.\n\nSo, we can rewrite:\n\\begin{align}\n\\frac{-x/3 + 1/6}{x^2 - x + 1} &= \\frac{-\\frac{1}{3}(x - 1/2)}{x^2 - x + 1}\\\\\n&= \\frac{-\\frac{1}{3} \\cdot \\frac{1}{2} \\cdot 2(x - 1/2)}{x^2 - x + 1}\\\\\n&= -\\frac{1}{6} \\cdot \\frac{2(x - 1/2)}{x^2 - x + 1}\\\\\n&= -\\frac{1}{6} \\cdot \\frac{\\frac{d}{dx}(x^2 - x + 1)}{x^2 - x + 1}\\\\\n&= -\\frac{1}{6} \\cdot \\frac{d}{dx}[\\ln(x^2 - x + 1)]\n\\end{align}\n\nThe integral of this is $-\\frac{1}{6}\\ln(x^2 - x + 1) + C_2$.\n\nFor $\\frac{1/2}{x^2 - x + 1}$, we can complete the square in the denominator:\n\\begin{align}\nx^2 - x + 1 &= x^2 - x + \\frac{1}{4} + 1 - \\frac{1}{4}\\\\\n&= (x - \\frac{1}{2})^2 + \\frac{3}{4}\n\\end{align}\n\nSo,\n\\begin{align}\n\\frac{1/2}{x^2 - x + 1} &= \\frac{1/2}{(x - \\frac{1}{2})^2 + \\frac{3}{4}}\\\\\n&= \\frac{1/2}{(x - \\frac{1}{2})^2 + (\\frac{\\sqrt{3}}{2})^2}\n\\end{align}\n\nWe can use the arctangent formula here: $\\int \\frac{1}{x^2 + a^2}dx = \\frac{1}{a}\\arctan(\\frac{x}{a}) + C$.\n\nApplying this to our integral:\n\\begin{align}\n\\int \\frac{1/2}{(x - \\frac{1}{2})^2 + (\\frac{\\sqrt{3}}{2})^2}dx &= \\frac{1/2}{\\frac{\\sqrt{3}}{2}}\\arctan\\left(\\frac{x - \\frac{1}{2}}{\\frac{\\sqrt{3}}{2}}\\right) + C_3\\\\\n&= \\frac{1}{\\sqrt{3}}\\arctan\\left(\\frac{2(x - \\frac{1}{2})}{\\sqrt{3}}\\right) + C_3\\\\\n&= \\frac{1}{\\sqrt{3}}\\arctan\\left(\\frac{2x - 1}{\\sqrt{3}}\\right) + C_3\n\\end{align}\n\nCombining all the pieces, the indefinite integral of $\\frac{1}{1+x^3}$ is:\n\\begin{align}\n\\int \\frac{1}{1+x^3}dx &= \\frac{1}{3}\\ln|x+1| - \\frac{1}{6}\\ln(x^2 - x + 1) + \\frac{1}{\\sqrt{3}}\\arctan\\left(\\frac{2x - 1}{\\sqrt{3}}\\right) + C\n\\end{align}\n\nOk actually I'm getting a bit confused with all the different terms. Let me try a different approach from the beginning.\n\nWe have the integral:\n\\begin{align}\n\\int \\frac{1}{1+x^3}dx\n\\end{align}\n\nWe know:\n\\begin{align}\n1 + x^3 &= (x+1)(x^2 - x + 1)\n\\end{align}\n\nNow, let's do a proper partial fraction decomposition:\n\\begin{align}\n\\frac{1}{1+x^3} &= \\frac{1}{(x+1)(x^2 - x + 1)}\\\\\n&= \\frac{A}{x+1} + \\frac{Bx + C}{x^2 - x + 1}\n\\end{align}\n\nWhen $x = -1$:\n\\begin{align}\n1 &= A \\cdot ((-1)^2 - (-1) + 1)\\\\\n&= A \\cdot (1 + 1 + 1)\\\\\n&= 3A\\\\\n\\Rightarrow A &= \\frac{1}{3}\n\\end{align}\n\nNow, let's find $B$ and $C$:\n\\begin{align}\n\\frac{1}{(x+1)(x^2 - x + 1)} &= \\frac{1/3}{x+1} + \\frac{Bx + C}{x^2 - x + 1}\n\\end{align}\n\nMultiply both sides by $(x+1)(x^2 - x + 1)$:\n\\begin{align}\n1 &= \\frac{1}{3}(x^2 - x + 1) + (Bx + C)(x+1)\\\\\n&= \\frac{1}{3}x^2 - \\frac{1}{3}x + \\frac{1}{3} + Bx^2 + Bx + Cx + C\n\\end{align}\n\nGrouping terms:\n\\begin{align}\n1 &= \\left(\\frac{1}{3} + B\\right)x^2 + \\left(-\\frac{1}{3} + B + C\\right)x + \\left(\\frac{1}{3} + C\\right)\n\\end{align}\n\nComparing coefficients:\n\\begin{align}\n\\frac{1}{3} + B &= 0\\\\\n-\\frac{1}{3} + B + C &= 0\\\\\n\\frac{1}{3} + C &= 1\n\\end{align}\n\nFrom the first equation, $B = -\\frac{1}{3}$.\nFrom the third equation, $C = \\frac{2}{3}$.\n\nSo:\n\\begin{align}\n\\frac{1}{1+x^3} &= \\frac{1/3}{x+1} + \\frac{-x/3 + 2/3}{x^2 - x + 1}\n\\end{align}\n\nLet's integrate the terms separately.\n\nFor the first term:\n\\begin{align}\n\\int \\frac{1/3}{x+1}dx &= \\frac{1}{3}\\ln|x+1| + C_1\n\\end{align}\n\nFor the second term, let's rewrite:\n\\begin{align}\n\\frac{-x/3 + 2/3}{x^2 - x + 1} &= \\frac{-x/3 + 1/6 + 1/2}{x^2 - x + 1}\n\\end{align}\n\nLet's consider:\n\\begin{align}\n\\frac{-x/3 + 1/6}{x^2 - x + 1} &= \\frac{-\\frac{1}{3}(x - 1/2)}{x^2 - x + 1}\n\\end{align}\n\nThe derivative of $x^2 - x + 1$ is $2x - 1 = 2(x - 1/2)$. So:\n\\begin{align}\n\\frac{-\\frac{1}{3}(x - 1/2)}{x^2 - x + 1} &= -\\frac{1}{6} \\cdot \\frac{2(x - 1/2)}{x^2 - x + 1}\\\\\n&= -\\frac{1}{6} \\cdot \\frac{d}{dx}[\\ln(x^2 - x + 1)]\n\\end{align}\n\nThe integral of this is:\n\\begin{align}\n\\int -\\frac{1}{6} \\cdot \\frac{d}{dx}[\\ln(x^2 - x + 1)]dx &= -\\frac{1}{6}\\ln(x^2 - x + 1) + C_2\n\\end{align}\n\nFor the term $\\frac{1/2}{x^2 - x + 1}$, we can complete the square in the denominator:\n\\begin{align}\nx^2 - x + 1 &= (x - \\frac{1}{2})^2 + \\frac{3}{4}\n\\end{align}\n\nSo:\n\\begin{align}\n\\frac{1/2}{x^2 - x + 1} &= \\frac{1/2}{(x - \\frac{1}{2})^2 + \\frac{3}{4}}\\\\\n&= \\frac{1/2}{(x - \\frac{1}{2})^2 + (\\frac{\\sqrt{3}}{2})^2}\n\\end{align}\n\nUsing the formula for the arctangent:\n\\begin{align}\n\\int \\frac{1/2}{(x - \\frac{1}{2})^2 + (\\frac{\\sqrt{3}}{2})^2}dx &= \\frac{1/2}{\\frac{\\sqrt{3}}{2}}\\arctan\\left(\\frac{x - \\frac{1}{2}}{\\frac{\\sqrt{3}}{2}}\\right) + C_3\\\\\n&= \\frac{1}{\\sqrt{3}}\\arctan\\left(\\frac{2(x - \\frac{1}{2})}{\\sqrt{3}}\\right) + C_3\\\\\n&= \\frac{1}{\\sqrt{3}}\\arctan\\left(\\frac{2x - 1}{\\sqrt{3}}\\right) + C_3\n\\end{align}\n\nCombining all the pieces, the indefinite integral of $\\frac{1}{1+x^3}$ is:\n\\begin{align}\n\\int \\frac{1}{1+x^3}dx &= \\frac{1}{3}\\ln|x+1| - \\frac{1}{6}\\ln(x^2 - x + 1) + \\frac{1}{\\sqrt{3}}\\arctan\\left(\\frac{2x - 1}{\\sqrt{3}}\\right) + C\n\\end{align}\n\nActually let's double-check because this still feels overly complicated.\n\nWe have:\n\\begin{align}\n\\frac{1}{1+x^3} &= \\frac{1/3}{x+1} + \\frac{-x/3 + 2/3}{x^2 - x + 1}\n\\end{align}\n\nThe first term's integral is straightforward:\n\\begin{align}\n\\int \\frac{1/3}{x+1}dx &= \\frac{1}{3}\\ln|x+1| + C_1\n\\end{align}\n\nLet's look closely at the second term:\n\\begin{align}\n\\frac{-x/3 + 2/3}{x^2 - x + 1}\n\\end{align}\n\nLet's create a substitution to simplify:\n\\begin{align}\nu &= x - \\frac{1}{2}\\\\\nx &= u + \\frac{1}{2}\\\\\ndx &= du\n\\end{align}\n\nThen:\n\\begin{align}\nx^2 - x + 1 &= (u + \\frac{1}{2})^2 - (u + \\frac{1}{2}) + 1\\\\\n&= u^2 + u + \\frac{1}{4} - u - \\frac{1}{2} + 1\\\\\n&= u^2 + \\frac{3}{4}\\\\\n&= u^2 + \\frac{\\sqrt{3}^2}{4}\n\\end{align}\n\nAnd:\n\\begin{align}\n-x/3 + 2/3 &= -(u + \\frac{1}{2})/3 + 2/3\\\\\n&= -u/3 - 1/6 + 2/3\\\\\n&= -u/3 + 1/2\n\\end{align}\n\nSo:\n\\begin{align}\n\\frac{-x/3 + 2/3}{x^2 - x + 1} &= \\frac{-u/3 + 1/2}{u^2 + \\frac{3}{4}}\n\\end{align}\n\nLet's split this into two fractions:\n\\begin{align}\n\\frac{-u/3 + 1/2}{u^2 + \\frac{3}{4}} &= \\frac{-u/3}{u^2 + \\frac{3}{4}} + \\frac{1/2}{u^2 + \\frac{3}{4}}\\\\\n&= -\\frac{1}{3} \\cdot \\frac{u}{u^2 + \\frac{3}{4}} + \\frac{1/2}{u^2 + \\frac{3}{4}}\n\\end{align}\n\nFor the first term, we know $\\int \\frac{u}{u^2 + a^2}du = \\frac{1}{2}\\ln(u^2 + a^2) + C$.\n\nSo:\n\\begin{align}\n\\int -\\frac{1}{3} \\cdot \\frac{u}{u^2 + \\frac{3}{4}}du &= -\\frac{1}{3} \\cdot \\frac{1}{2}\\ln(u^2 + \\frac{3}{4}) + C_2\\\\\n&= -\\frac{1}{6}\\ln(u^2 + \\frac{3}{4}) + C_2\\\\\n&= -\\frac{1}{6}\\ln(x^2 - x + 1) + C_2\n\\end{align}\n\nFor the second term, we use the arctangent formula:\n\\begin{align}\n\\int \\frac{1/2}{u^2 + \\frac{3}{4}}du &= \\frac{1/2}{\\sqrt{\\frac{3}{4}}}\\arctan\\left(\\frac{u}{\\sqrt{\\frac{3}{4}}}\\right) + C_3\\\\\n&= \\frac{1/2}{\\frac{\\sqrt{3}}{2}}\\arctan\\left(\\frac{u}{\\frac{\\sqrt{3}}{2}}\\right) + C_3\\\\\n&= \\frac{1}{\\sqrt{3}}\\arctan\\left(\\frac{2u}{\\sqrt{3}}\\right) + C_3\\\\\n&= \\frac{1}{\\sqrt{3}}\\arctan\\left(\\frac{2(x - \\frac{1}{2})}{\\sqrt{3}}\\right) + C_3\\\\\n&= \\frac{1}{\\sqrt{3}}\\arctan\\left(\\frac{2x - 1}{\\sqrt{3}}\\right) + C_3\n\\end{align}\n\nCombining all the pieces, the indefinite integral of $\\frac{1}{1+x^3}$ is:\n\\begin{align}\n\\int \\frac{1}{1+x^3}dx &= \\frac{1}{3}\\ln|x+1| - \\frac{1}{6}\\ln(x^2 - x + 1) + \\frac{1}{\\sqrt{3}}\\arctan\\left(\\frac{2x - 1}{\\sqrt{3}}\\right) + C\n\\end{align}\n\nOk now we have verified the integral, let's check once more.\n\nActually, let's use another method to check. We'll perform a direct substitution:\n\nLet $u = x - \\frac{1}{2}$, then $x = u + \\frac{1}{2}$ and $dx = du$.\n\n\\begin{align}\n\\int \\frac{1}{1+x^3}dx &= \\int \\frac{1}{1+(u + \\frac{1}{2})^3}du\\\\\n&= \\int \\frac{1}{1+u^3 + \\frac{3}{2}u^2 + \\frac{3}{4}u + \\frac{1}{8}}du\n\\end{align}\n\nThis doesn't simplify much. Let's stick with the original approach and the result we've already derived.\n\nLet's go ahead and reconfirm our result. We have:\n\\begin{align}\n\\int \\frac{1}{1+x^3}dx &= \\frac{1}{3}\\ln|x+1| - \\frac{1}{6}\\ln(x^2 - x + 1) + \\frac{1}{\\sqrt{3}}\\arctan\\left(\\frac{2x - 1}{\\sqrt{3}}\\right) + C\n\\end{align}\n\nIn fact this does check out. In summary, the indefinite integral of $\\frac{1}{1+x^3}$ is:\n\\begin{align}\n\\int \\frac{1}{1+x^3}dx &= \\frac{1}{3}\\ln|x+1| - \\frac{1}{6}\\ln(x^2 - x + 1) + \\frac{1}{\\sqrt{3}}\\arctan\\left(\\frac{2x - 1}{\\sqrt{3}}\\right) + C\n\\end{align}\n\nWhere $C$ is the constant of integration.",
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"text": "# 不定积分 $\\int \\frac{1}{1+x^3} dx$\n\n首先我需要计算 $\\frac{1}{1+x^3}$ 的不定积分。这个积分需要使用部分分式分解法。\n\n## 部分分式分解\n\n第一步，对 $1+x^3$ 进行因式分解：\n$$1+x^3 = (x+1)(x^2-x+1)$$\n\n然后进行部分分式分解：\n$$\\frac{1}{1+x^3} = \\frac{A}{x+1} + \\frac{Bx+C}{x^2-x+1}$$\n\n当 $x = -1$ 时：\n$$1 = A((-1)^2-(-1)+1) = 3A \\Rightarrow A = \\frac{1}{3}$$\n\n对比系数法求解 $B$ 和 $C$：\n$$\\frac{1}{3}(x^2-x+1) + (Bx+C)(x+1) = 1$$\n\n展开并比较系数：\n$$\\frac{1}{3}x^2 - \\frac{1}{3}x + \\frac{1}{3} + Bx^2 + Bx + Cx + C = 1$$\n\n得到：\n$$\\left\\{\\begin{array}{l}\n\\frac{1}{3}+B=0 \\\\\n-\\frac{1}{3}+B+C=0 \\\\\n\\frac{1}{3}+C=1\n\\end{array}\\right.$$\n\n解得：$B = -\\frac{1}{3}$，$C = \\frac{2}{3}$。\n\n因此：\n$$\\frac{1}{1+x^3} = \\frac{1/3}{x+1} + \\frac{-x/3+2/3}{x^2-x+1}$$\n\n## 积分计算\n\n1. 对于第一项： $\\int \\frac{1/3}{x+1}dx = \\frac{1}{3}\\ln|x+1|$\n\n2. 对于第二项，将 $\\frac{-x/3+2/3}{x^2-x+1}$ 重写为：\n $$\\frac{-x/3+2/3}{x^2-x+1} = -\\frac{1}{6} \\cdot \\frac{d}{dx}[\\ln(x^2-x+1)] + \\frac{1/2}{x^2-x+1}$$\n\n (这里利用了 $\\frac{d}{dx}(x^2-x+1) = 2x-1$ 和部分分式调整)\n\n • $\\int -\\frac{1}{6} \\cdot \\frac{d}{dx}[\\ln(x^2-x+1)]dx = -\\frac{1}{6}\\ln(x^2-x+1)$\n \n • 对于 $\\frac{1/2}{x^2-x+1}$，注意到 $x^2-x+1 = (x-\\frac{1}{2})^2+\\frac{3}{4}$，所以：\n $$\\int \\frac{1/2}{x^2-x+1}dx = \\frac{1}{\\sqrt{3}}\\arctan\\left(\\frac{2x-1}{\\sqrt{3}}\\right)$$\n\n结合所有部分，最终积分为：\n$$\\int \\frac{1}{1+x^3}dx = \\frac{1}{3}\\ln|x+1| - \\frac{1}{6}\\ln(x^2-x+1) + \\frac{1}{\\sqrt{3}}\\arctan\\left(\\frac{2x-1}{\\sqrt{3}}\\right) + C$$\n\n其中 $C$ 为积分常数。"
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}2.openai格式接入claude-3-7-sonnet-20250219-thinking
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如果用户没传参就适配，传参了就不做更改
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}4.newapi下游接入，并仍支持上述方法
更新至calciumion/new-api-horizon:latest（0.1.2.9.4及以后版本）接入时选择Anthropic Claude方式接入，填入两个模型名称即可
imagen-3.0-generate-002
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价格 | Generative AI on Vertex AI | Google Cloud
api接入chatluna
准备:已使用koishi接入QQ
QQ群282381753
安装chatluna和chatluna-openai-like-adapter
api接口填https://api.bailili.top/v1
api接入sus-chat
准备:已使用koishi接入QQ
QQ群775791760
安装sus-chat
api接口填https://api.bailili.top/v1
api接入酒馆
推荐edge浏览器进入，禁止用QQ进入
连接中转api：
上传人格：
进入QQ群的群文件，下载：早期破限组角色卡分类包.zip
解压密码wangyue520，解压到一个找得到的地方。
进入网站，这里导入想玩的人格文件，格式就是png直接导入
点右上角角色卡，选择角色即可对话
如果是手机，选完角色会出现下图，点左下角（或右下角）缝隙处开始对话
claude模型验证
claude作为模型掺假的重灾区，各种逆向掺假的低价渠道层出不穷，逆向会出现文本截断，其他来源不明的渠道回复质量更无法保证
本站坚持并保证使用官方api，保证回复质量，可自行进行对话验证和温度验证
对话验证
区分claude3-7官方/逆向方法：
询问对方是claude几-几？逆向会说自己是3-7，官api会说自己是3-opus（推测3-7为3-opus升级而来）
温度验证
模型鉴定网站：API CHECK（防君子不防小人）
使用以上网站进行claude-3-7-sonnet-20250219温度验证，本站符合官方key参考响应：
如果数字不是6则为其他低价渠道，可能是逆向或者掺假，来源不明，模型质量无法保证：
某低价中转站：
其他模型官方返回结果参考：
claude-3-5-sonnet-20241022以前是82现在是91
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