| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the indefinite integral of the function \(\frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}}\), we will use the substitution method. Let's proceed step by step. 1. **Substitution**: Let \( u = \sqrt{4x + 2} \). Then, we need to express \( x \) in terms of \( u \): \[ u = \sqrt{4x + 2} \implies u^2 = 4x + 2 \implies 4x = u^2 - 2 \implies x = \frac{u^2 - 2}{4} \] 2. **Differentiate \( u \) with respect to \( x \)**: \[ \frac{du}{dx} = \frac{d}{dx} \left( \sqrt{4x + 2} \right) = \frac{1}{2\sqrt{4x + 2}} \cdot 4 = \frac{2}{\sqrt{4x + 2}} \] Therefore, \[ du = \frac{2}{\sqrt{4x + 2}} \, dx \implies dx = \frac{\sqrt{4x + 2}}{2} \, du = \frac{u}{2} \, du \] 3. **Rewrite the integral in terms of \( u \)**: \[ \int \frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}} \, dx = \int \frac{2 e^u}{u} \cdot \frac{u}{2} \, du = \int e^u \, du \] 4. **Integrate**: \[ \int e^u \, du = e^u + C \] 5. **Substitute back \( u = \sqrt{4x + 2} \)**: \[ e^u + C = e^{\sqrt{4x + 2}} + C \] Thus, the indefinite integral is: \[ \boxed{e^{\sqrt{4x + 2}} + C} \] |
e^{\sqrt{4 x+2}} |
| deepseek-r1 | To find the indefinite integral of the function \(\frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}}\), we can use substitution. 1. Let \(u = \sqrt{4x + 2}\). 2. Compute the derivative of \(u\) with respect to \(x\): \[ \frac{du}{dx} = \frac{1}{2\sqrt{4x + 2}} \cdot 4 = \frac{2}{\sqrt{4x + 2}} \] 3. Rearrange to express \(du\) in terms of \(dx\): \[ du = \frac{2}{\sqrt{4x + 2}} \, dx \] 4. Substitute \(u\) and \(du\) into the integral: \[ \int \frac{2 e^{\sqrt{4x + 2}}}{\sqrt{4x + 2}} \, dx = \int e^u \, du \] 5. Integrate with respect to \(u\): \[ \int e^u \, du = e^u + C \] 6. Substitute back \(u = \sqrt{4x + 2}\): \[ e^{\sqrt{4x + 2}} + C \] Thus, the indefinite integral is \(\boxed{e^{\sqrt{4x + 2}} + C}\). |
e^{\sqrt{4 x+2}} |
| gpto1 | To solve the integral: \[ \int \frac{2 e^{\sqrt{4x+2}}}{\sqrt{4x+2}} \, dx \] We'll use a substitution to simplify the expression. Let: \[ u = \sqrt{4x+2} \] Compute \( du \): \[ u = (4x + 2)^{1/2} \\ du = \frac{1}{2}(4x + 2)^{-1/2} \cdot 4 \, dx = \frac{2}{\sqrt{4x + 2}} \, dx \] Rewriting \( du \): \[ du = \frac{2}{\sqrt{4x+2}} \, dx \\ \Rightarrow \frac{2}{\sqrt{4x+2}} \, dx = du \] This matches the numerator and \( dx \) in the original integral. Substitute back into the integral: \[ \int e^{\sqrt{4x+2}} \cdot \frac{2}{\sqrt{4x+2}} \, dx = \int e^{u} \, du \] Integrate \( e^u \): \[ \int e^{u} \, du = e^{u} + C \] Substitute back \( u = \sqrt{4x+2} \): \[ e^{\sqrt{4x+2}} + C \] **Answer:** \boxed{\,e^{\,\sqrt{4x+2}}\,+\,C\,} |
e^{\sqrt{4 x+2}} |
| gpt-4o-2024-08-06 | To find the indefinite integral of the function \(\frac{2 e^{\sqrt{4x+2}}}{\sqrt{4x+2}}\), we can use a substitution method. Let's set: \[ u = \sqrt{4x + 2} \] Then, we have: \[ u^2 = 4x + 2 \] Differentiating both sides with respect to \(x\), we get: \[ 2u \frac{du}{dx} = 4 \quad \Rightarrow \quad \frac{du}{dx} = \frac{2}{u} \quad \Rightarrow \quad dx = \frac{u}{2} \, du \] Substituting these into the integral, we have: \[ \int \frac{2 e^{\sqrt{4x+2}}}{\sqrt{4x+2}} \, dx = \int \frac{2 e^u}{u} \cdot \frac{u}{2} \, du \] Simplifying the expression inside the integral: \[ = \int e^u \, du \] The integral of \(e^u\) with respect to \(u\) is simply \(e^u\). Therefore, we have: \[ \int e^u \, du = e^u + C \] Substituting back \(u = \sqrt{4x + 2}\), we obtain: \[ e^u = e^{\sqrt{4x + 2}} \] Thus, the indefinite integral is: \[ e^{\sqrt{4x + 2}} + C \] Therefore, the final answer is: \[ \boxed{e^{\sqrt{4x + 2}} + C} \] |
e^{\sqrt{4 x+2}} |
| DeepSeek-Coder-V2-Lite-Instruct | To find the indefinite integral of the function \(\frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}}\), we can use the method of substitution. Let's proceed step-by-step: 1. **Substitution**: Let \( u = \sqrt{4 x + 2} \). 2. **Differentiate \( u \) with respect to \( x \)**: \[ u = \sqrt{4 x + 2} \implies u^2 = 4 x + 2 \implies 2 u \, du = 4 \, dx \implies dx = \frac{u \, du}{2} \] 3. **Rewrite the integral in terms of \( u \)**: \[ \int \frac{2 e^u}{u} \cdot \frac{u \, du}{2} = \int e^u \, du \] 4. **Integrate \( e^u \)**: \[ \int e^u \, du = e^u + C \] 5. **Substitute back \( u = \sqrt{4 x + 2} \)**: \[ e^u = e^{\sqrt{4 x + 2}} \] 6. **Combine the results**: \[ \int \frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}} \, dx = e^{\sqrt{4 x + 2}} + C \] Thus, the indefinite integral of the given function is: \[ \boxed{e^{\sqrt{4 x + 2}} + C} \] |
e^{\sqrt{4 x+2}} |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the integral to be solved The integral to be solved is $\int \frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}} dx$. ## Step 2: Choose a suitable substitution Let's use the substitution $u = \sqrt{4x + 2}$, which implies $du = \frac{4}{2\sqrt{4x + 2}} dx = \frac{2}{\sqrt{4x + 2}} dx$. ## Step 3: Express the integral in terms of the new variable The integral becomes $\int 2 e^u \cdot \frac{1}{2} du = \int e^u du$. ## Step 4: Solve the integral in terms of the new variable The integral of $e^u$ with respect to $u$ is $e^u + C$. ## Step 5: Substitute back to the original variable Substituting back $u = \sqrt{4x + 2}$, we get the solution as $e^{\sqrt{4x + 2}} + C$. The final answer is: $\boxed{e^{\sqrt{4x + 2}}}$ |
e^{\sqrt{4 x+2}} |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Identify the integral to be solved We need to find the indefinite integral of the function $\frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}}$. ## Step 2: Choose a suitable substitution Let's use the substitution $u = \sqrt{4x + 2}$. This will simplify the integral. ## Step 3: Find the derivative of u with respect to x The derivative of $u$ with respect to $x$ is $\frac{du}{dx} = \frac{2}{\sqrt{4x + 2}} = \frac{2}{u}$. ## Step 4: Express du in terms of dx Rearranging the equation from step 3, we get $du = \frac{2}{u} dx$. ## Step 5: Rewrite the integral in terms of u Substituting $u = \sqrt{4x + 2}$ and $du = \frac{2}{u} dx$ into the original integral, we get $\int \frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}} dx = \int \frac{2 e^{u}}{u} \cdot \frac{u}{2} du = \int e^{u} du$. ## Step 6: Evaluate the integral The integral of $e^{u}$ with respect to $u$ is $e^{u} + C$, where $C$ is the constant of integration. ## Step 7: Substitute back to the original variable Substituting back $u = \sqrt{4x + 2}$, we get the final answer as $e^{\sqrt{4x + 2}} + C$. The final answer is: $\boxed{e^{\sqrt{4x + 2}}}$ |
e^{\sqrt{4 x+2}} |
| Qwen2-72B-Instruct | To find the indefinite integral of the given function, we will use substitution to simplify the expression. Let's set: $$ u = \sqrt{4x + 2} $$ Then, $$ du = \frac{1}{2\sqrt{4x + 2}} \cdot 4 dx = \frac{2}{\sqrt{4x + 2}} dx $$ So, $$ dx = \frac{\sqrt{4x + 2}}{2} du = \frac{u}{2} du $$ Now, we can rewrite the original integral in terms of \( u \): $$ \int \frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}} dx = \int 2e^u \cdot \frac{u}{2} du = \int ue^u du $$ This is now a standard integral that can be solved using integration by parts. Let's set: $$ v = u, dv = du $$ $$ w = e^u, dw = e^u du $$ Integration by parts formula is: $$ \int vdw = vw - \int wdv $$ Applying this formula: $$ \int ue^u du = ue^u - \int e^u du = ue^u - e^u + C $$ Substituting back for \( u \): $$ \int \frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}} dx = (\sqrt{4x + 2})e^{\sqrt{4x + 2}} - e^{\sqrt{4x + 2}} + C $$ Simplifying: $$ = e^{\sqrt{4x + 2}}(\sqrt{4x + 2} - 1) + C $$ Therefore, the indefinite integral of the given function is: $$ \boxed{e^{\sqrt{4x + 2}}(\sqrt{4x + 2} - 1) + C} $$ |
e^{\sqrt{4 x+2}} |
| baichuan4 | 为了找到给定函数的不定积分，我们可以使用换元积分法。考虑到函数的形式，我们可以设 $u = \sqrt{4x + 2}$，然后对两边求导得到 $du = \frac{2}{\sqrt{4x + 2}}dx$。现在我们可以将原函数中的 $\frac{2}{\sqrt{4x + 2}}dx$ 替换为 $du$，并且将 $e^{\sqrt{4x + 2}}$ 替换为 $e^u$。 原函数变为： $$\int \frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}} dx = \int e^u du$$ 这是一个基本的指数函数积分，其积分结果是 $e^u + C$，其中 $C$ 是积分常数。 现在我们需要将 $u$ 替换回原来的变量 $x$。由于我们设定了 $u = \sqrt{4x + 2}$，我们可以直接将 $u$ 替换回去得到最终答案： $$\int \frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}} dx = e^{\sqrt{4x + 2}} + C$$ 因此，给定函数的不定积分是： $$\boxed{e^{\sqrt{4x + 2}} + C}$$ 这里的 $C$ 是积分常数，表示积分的结果可以加上一个任意常数。 |
e^{\sqrt{4 x+2}} |
| doubao | Let $u=\sqrt{4x+2}$, then $u^2=4x+2$, $2udu=4dx$, $dx=\frac{u}{2}du$. So the integral becomes: \[ \begin{align*} &\int\frac{2e^{\sqrt{4x+2}}}{\sqrt{4x+2}}dx\\ =&\int\frac{2e^u}{u}\cdot\frac{u}{2}du\\ =&\int e^udu\\ =&e^u+C\\ =&\boxed{e^{\sqrt{4x+2}}+C} \end{align*} \] |
e^{\sqrt{4 x+2}} |
| gpto1 | To solve the integral: \[ \int \frac{2 e^{\sqrt{4x+2}}}{\sqrt{4x+2}} \, dx \] We'll use a substitution to simplify the expression. Let: \[ u = \sqrt{4x+2} \] Compute \( du \): \[ u = (4x + 2)^{1/2} \\ du = \frac{1}{2}(4x + 2)^{-1/2} \cdot 4 \, dx = \frac{2}{\sqrt{4x + 2}} \, dx \] Rewriting \( du \): \[ du = \frac{2}{\sqrt{4x+2}} \, dx \\ \Rightarrow \frac{2}{\sqrt{4x+2}} \, dx = du \] This matches the numerator and \( dx \) in the original integral. Substitute back into the integral: \[ \int e^{\sqrt{4x+2}} \cdot \frac{2}{\sqrt{4x+2}} \, dx = \int e^{u} \, du \] Integrate \( e^u \): \[ \int e^{u} \, du = e^{u} + C \] Substitute back \( u = \sqrt{4x+2} \): \[ e^{\sqrt{4x+2}} + C \] **Answer:** \boxed{\,e^{\,\sqrt{4x+2}}\,+\,C\,} |
e^{\sqrt{4 x+2}} |
| kimi | To find the indefinite integral of the function \(\frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}}\), we will use substitution. Let's proceed step by step. 1. **Substitution:** Let \( u = \sqrt{4x + 2} \). 2. **Differentiate \( u \):** \[ u = \sqrt{4x + 2} \] Squaring both sides, we get: \[ u^2 = 4x + 2 \] Differentiate both sides with respect to \( x \): \[ 2u \, du = 4 \, dx \] Solving for \( dx \): \[ dx = \frac{u}{2} \, du \] 3. **Rewrite the integral in terms of \( u \):** Substitute \( u \) and \( dx \) into the original integral: \[ \int \frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}} \, dx = \int \frac{2 e^u}{u} \cdot \frac{u}{2} \, du \] Simplify the integrand: \[ = \int e^u \, du \] 4. **Integrate with respect to \( u \):** The integral of \( e^u \) is \( e^u \): \[ \int e^u \, du = e^u + C \] 5. **Substitute back \( u = \sqrt{4x + 2} \):** \[ e^u + C = e^{\sqrt{4x + 2}} + C \] Therefore, the indefinite integral of the given function is: \[ \boxed{e^{\sqrt{4x + 2}} + C} \] |
e^{\sqrt{4 x+2}} |
| minimax | To find the indefinite integral of the function $\frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}}$, we can use a substitution method. Let's set $u = \sqrt{4x + 2}$, which implies that $u^2 = 4x + 2$. Differentiating both sides with respect to $x$ gives us $2u \frac{du}{dx} = 4$, or $\frac{du}{dx} = \frac{2}{u}$. Now, we can rewrite the integral in terms of $u$: $$\int \frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}} dx = \int \frac{2 e^u}{u} \cdot \frac{du}{dx} dx = \int \frac{2 e^u}{u} \cdot \frac{2}{u} du = \int \frac{4 e^u}{u^2} du.$$ Now we have an integral of the form $\int \frac{f'(u)}{f(u)} du$, which is a standard integral that can be solved by recognizing it as the |
e^{\sqrt{4 x+2}} |
| wenxinyiyan | 首先，我们观察给定的函数 $\frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}}$，并尝试通过换元法来简化它。 令 $u = \sqrt{4 x+2}$，则 $u^2 = 4x + 2$，从而 $x = \frac{u^2 - 2}{4}$。 对 $u$ 求导，得到 $du = \frac{1}{2} \times (4x + 2)^{-\frac{1}{2}} \times 4 dx = \frac{2}{\sqrt{4x+2}} dx$。 将 $dx$ 表达为 $du$ 的形式：$dx = \frac{\sqrt{4x+2}}{2} du = \frac{u}{2} du$。 将 $u$ 和 $du$ 代入原函数，得到： $\frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}} dx = \frac{2 e^u}{u} \times \frac{u}{2} du = e^u du$ 对 $e^u du$ 进行积分，得到： $\int e^u du = e^u + C$ 将 $u = \sqrt{4x+2}$ 代回，得到： $e^u + C = e^{\sqrt{4x+2}} + C$ 所以，最终的不定积分为： $\boxed{e^{\sqrt{4x+2}} + C}$ |
e^{\sqrt{4 x+2}} |
| zhipu | To find the indefinite integral of the function \( \frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}} \), we can use a substitution method. Let \( u = \sqrt{4x + 2} \). Then: \[ u^2 = 4x + 2 \] Differentiating both sides with respect to \( x \): \[ 2u \frac{du}{dx} = 4 \] \[ \frac{du}{dx} = \frac{4}{2u} = \frac{2}{u} \] \[ dx = \frac{u}{2} du \] Now, substitute \( u = \sqrt{4x + 2} \) and \( dx = \frac{u}{2} du \) into the integral: \[ \int \frac{2 e^{\sqrt{4 x+2}}}{\sqrt{4 x+2}} \, dx = \int \frac{2 e^u}{u} \cdot \frac{u}{2} \, du \] Simplify the integrand: \[ \int \frac{2 e^u}{u} \cdot \frac{u}{2} \, du = \int e^u \, du \] The integral of \( e^u \) with respect to \( u \) is simply \( e^u \): \[ \int e^u \, du = e^u + C \] Finally, substitute back \( u = \sqrt{4x + 2} \): \[ e^u + C = e^{\sqrt{4x + 2}} + C \] Thus, the indefinite integral is: \[ \boxed{e^{\sqrt{4x + 2}} + C} \] |
e^{\sqrt{4 x+2}} |
| deepseek-v3 | To find the indefinite integral of the function \(\frac{2 e^{\sqrt{4x+2}}}{\sqrt{4x+2}}\), we can proceed with the following steps: 1. **Substitution:** Let \( u = \sqrt{4x + 2} \). Then, we can find the differential \( du \) in terms of \( dx \): \[ u = \sqrt{4x + 2} \implies u^2 = 4x + 2 \] Differentiating both sides: \[ 2u \, du = 4 \, dx \implies dx = \frac{u}{2} \, du \] 2. **Rewrite the Integral:** Substitute \( u \) and \( dx \) into the original integral: \[ \int \frac{2 e^{u}}{u} \cdot \frac{u}{2} \, du = \int e^{u} \, du \] Notice that the \( u \) terms cancel out: \[ \int e^{u} \, du = e^{u} + C \] 3. **Back-Substitution:** Replace \( u \) with the original expression in terms of \( x \): \[ e^{u} + C = e^{\sqrt{4x + 2}} + C \] Therefore, the indefinite integral is: \[ \boxed{e^{\sqrt{4x + 2}} + C} \] |
e^{\sqrt{4 x+2}} |