| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the indefinite integral of the function \(\frac{4}{3 \sqrt[3]{4 x-\frac{17}{2}}^2}\), we will follow a step-by-step approach. First, let's rewrite the function in a more manageable form: \[ \frac{4}{3 \sqrt[3]{4 x-\frac{17}{2}}^2} = \frac{4}{3} \cdot (4x - \frac{17}{2})^{-2/3} \] Next, we will use the substitution method to simplify the integral. Let: \[ u = 4x - \frac{17}{2} \] Then, the differential \(du\) is: \[ du = 4 \, dx \implies dx = \frac{du}{4} \] Now, substitute \(u\) and \(dx\) into the integral: \[ \int \frac{4}{3} \cdot (4x - \frac{17}{2})^{-2/3} \, dx = \int \frac{4}{3} \cdot u^{-2/3} \cdot \frac{du}{4} \] Simplify the integral: \[ = \frac{4}{3} \cdot \frac{1}{4} \int u^{-2/3} \, du \] \[ = \frac{1}{3} \int u^{-2/3} \, du \] Now, integrate \(u^{-2/3}\): \[ \int u^{-2/3} \, du = \frac{u^{-2/3 + 1}}{-2/3 + 1} = \frac{u^{1/3}}{1/3} = 3u^{1/3} \] So, the integral becomes: \[ \frac{1}{3} \cdot 3u^{1/3} = u^{1/3} \] Substitute back \(u = 4x - \frac{17}{2}\): \[ u^{1/3} = (4x - \frac{17}{2})^{1/3} \] Therefore, the indefinite integral is: \[ \int \frac{4}{3 \sqrt[3]{4 x-\frac{17}{2}}^2} \, dx = (4x - \frac{17}{2})^{1/3} + C \] Putting the final answer in a box: \[ \boxed{(4x - \frac{17}{2})^{1/3} + C} \] |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |
| deepseek-r1 | To find the indefinite integral of the function \(\frac{4}{3 \sqrt[3]{4 x - \frac{17}{2}}^2}\), we start by rewriting the function in a more convenient form for integration: \[ \frac{4}{3 \left(4x - \frac{17}{2}\right)^{\frac{2}{3}}} \] We use the substitution \(u = 4x - \frac{17}{2}\). Then, the derivative of \(u\) with respect to \(x\) is: \[ \frac{du}{dx} = 4 \quad \Rightarrow \quad du = 4 \, dx \quad \Rightarrow \quad dx = \frac{du}{4} \] Substituting \(u\) and \(dx\) into the integral, we get: \[ \int \frac{4}{3} u^{-\frac{2}{3}} \cdot \frac{du}{4} = \frac{1}{3} \int u^{-\frac{2}{3}} \, du \] Integrating \(u^{-\frac{2}{3}}\) gives: \[ \frac{1}{3} \cdot \frac{u^{\frac{1}{3}}}{\frac{1}{3}} + C = u^{\frac{1}{3}} + C \] Substituting back \(u = 4x - \frac{17}{2}\), we obtain: \[ \left(4x - \frac{17}{2}\right)^{\frac{1}{3}} + C \] This can be written using a cube root as: \[ \sqrt[3]{4x - \frac{17}{2}} + C \] Thus, the indefinite integral is: \[ \boxed{\sqrt[3]{4x - \frac{17}{2}} + C} \] |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |
| gpto1 | To find the indefinite integral of the function \(\frac{4}{3 \left( \sqrt[3]{4x - \dfrac{17}{2}} \right)^2}\), we can rewrite the integral in terms of exponents and then apply substitution. First, express the function with exponents: \[ \int \frac{4}{3 \left( \sqrt[3]{4x - \dfrac{17}{2}} \right)^2} \, dx = \frac{4}{3} \int \left(4x - \dfrac{17}{2}\right)^{-2/3} \, dx \] Now, let us make a substitution: Let \( u = 4x - \dfrac{17}{2} \), so \( du = 4 \, dx \) or \( dx = \dfrac{du}{4} \). Substitute into the integral: \[ \frac{4}{3} \int u^{-2/3} \cdot \frac{du}{4} = \frac{1}{3} \int u^{-2/3} \, du \] Now, integrate \( u^{-2/3} \): \[ \int u^{-2/3} \, du = \frac{ u^{1/3} }{1/3} + C = 3u^{1/3} + C \] Multiply by \(\dfrac{1}{3}\): \[ \frac{1}{3} \cdot 3u^{1/3} = u^{1/3} \] Replace \( u = 4x - \dfrac{17}{2} \) and express back in radical form: \[ u^{1/3} = \left( 4x - \dfrac{17}{2} \right)^{1/3} = \sqrt[3]{4x - \dfrac{17}{2}} \] Thus, the indefinite integral is: \[ \int \frac{4}{3 \left( \sqrt[3]{4x - \dfrac{17}{2}} \right)^2} \, dx = \sqrt[3]{4x - \dfrac{17}{2}} + C \] \(\boxed{\,\sqrt[3]{4x\ -\ \dfrac{17}{2}}\ +\ C\,}\) |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |
| gpt-4o-2024-08-06 | To find the indefinite integral of the function \(\frac{4}{3 \sqrt[3]{4x - \frac{17}{2}}^2}\), we start by rewriting the integrand in a more convenient form for integration. The expression \(\sqrt[3]{4x - \frac{17}{2}}\) can be written as \((4x - \frac{17}{2})^{1/3}\). Therefore, the integrand becomes: \[ \frac{4}{3 (4x - \frac{17}{2})^{2/3}} \] This can be rewritten as: \[ \frac{4}{3} (4x - \frac{17}{2})^{-2/3} \] We will use the substitution method to solve this integral. Let: \[ u = 4x - \frac{17}{2} \] Then, the derivative of \(u\) with respect to \(x\) is: \[ \frac{du}{dx} = 4 \quad \Rightarrow \quad dx = \frac{du}{4} \] Substituting \(u\) and \(dx\) into the integral, we have: \[ \int \frac{4}{3} u^{-2/3} \cdot \frac{du}{4} \] Simplifying, this becomes: \[ \frac{1}{3} \int u^{-2/3} \, du \] Now, we integrate \(u^{-2/3}\). The antiderivative of \(u^n\) is \(\frac{u^{n+1}}{n+1}\), provided \(n \neq -1\). Here, \(n = -\frac{2}{3}\), so: \[ \int u^{-2/3} \, du = \frac{u^{-2/3 + 1}}{-2/3 + 1} = \frac{u^{1/3}}{1/3} = 3u^{1/3} \] Thus, the integral becomes: \[ \frac{1}{3} \cdot 3u^{1/3} = u^{1/3} \] Substituting back \(u = 4x - \frac{17}{2}\), we have: \[ (4x - \frac{17}{2})^{1/3} \] Therefore, the indefinite integral of the given function is: \[ \boxed{(4x - \frac{17}{2})^{1/3} + C} \] where \(C\) is the constant of integration. |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |
| DeepSeek-Coder-V2-Lite-Instruct | To find the indefinite integral of the function \(\frac{4}{3 \sqrt[3]{4 x-\frac{17}{2}}^2}\), we will follow these steps: 1. **Substitution**: Let \( u = 4x - \frac{17}{2} \). Then, \( du = 4 \, dx \), which implies \( dx = \frac{du}{4} \). 2. **Rewrite the integral**: Substitute \( u \) and \( dx \) into the original integral: \[ \int \frac{4}{3 \sqrt[3]{u}^2} \, dx = \int \frac{4}{3 u^{2/3}} \cdot \frac{du}{4} \] 3. **Simplify the integral**: The constants \( 4 \) and \( \frac{1}{4} \) cancel out: \[ \int \frac{4}{3 u^{2/3}} \cdot \frac{du}{4} = \int \frac{1}{3} u^{-2/3} \, du \] 4. **Integrate**: Use the power rule for integration, which states \(\int u^n \, du = \frac{u^{n+1}}{n+1} + C\), where \( n \neq -1 \): \[ \int \frac{1}{3} u^{-2/3} \, du = \frac{1}{3} \cdot \frac{u^{-2/3 + 1}}{-2/3 + 1} + C \] \[ = \frac{1}{3} \cdot \frac{u^{1/3}}{1/3} + C \] \[ = \frac{1}{3} \cdot 3 u^{1/3} + C \] \[ = u^{1/3} + C \] 5. **Substitute back**: Replace \( u \) with \( 4x - \frac{17}{2} \): \[ u^{1/3} + C = \left(4x - \frac{17}{2}\right)^{1/3} + C \] Thus, the indefinite integral of the given function is: \[ \boxed{\left(4x - \frac{17}{2}\right)^{1/3} + C} \] |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the integral to be solved The given integral is $\int \frac{4}{3 \sqrt[3]{4 x-\frac{17}{2}}^2} dx$. ## Step 2: Simplify the integral expression Let's simplify the expression inside the integral. We can rewrite $\sqrt[3]{4 x-\frac{17}{2}}^2$ as $(4 x-\frac{17}{2})^{\frac{2}{3}}$. ## Step 3: Substitute to simplify the integral To simplify the integral, we can substitute $u = 4x - \frac{17}{2}$. Then, $du = 4 dx$. ## Step 4: Express the integral in terms of u The integral becomes $\int \frac{4}{3 u^{\frac{2}{3}}} \frac{1}{4} du = \frac{1}{3} \int u^{-\frac{2}{3}} du$. ## Step 5: Solve the integral The integral of $u^{-\frac{2}{3}}$ with respect to $u$ is $\frac{u^{\frac{1}{3}}}{\frac{1}{3}} = 3u^{\frac{1}{3}}$. ## Step 6: Substitute back to the original variable Substituting back $u = 4x - \frac{17}{2}$, we get $\frac{1}{3} \cdot 3(4x - \frac{17}{2})^{\frac{1}{3}} = (4x - \frac{17}{2})^{\frac{1}{3}}$. ## Step 7: Add the constant of integration The indefinite integral is $(4x - \frac{17}{2})^{\frac{1}{3}} + C$. The final answer is: $\boxed{(4x - \frac{17}{2})^{\frac{1}{3}} + C}$ |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^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 $\frac{4}{3 \sqrt[3]{4 x-\frac{17}{2}}^2}$. ## Step 2: Rewrite the integral in a more manageable form Let's rewrite the integral as $\frac{4}{3} \int \frac{1}{\sqrt[3]{4 x-\frac{17}{2}}^2} dx$. ## Step 3: Make a substitution to simplify the integral To simplify the integral, let's make the substitution $u = 4x - \frac{17}{2}$. Then $du = 4dx$, or $dx = \frac{1}{4}du$. ## Step 4: Rewrite the integral in terms of the new variable The integral becomes $\frac{4}{3} \int \frac{1}{u^{\frac{2}{3}}} \cdot \frac{1}{4} du = \frac{1}{3} \int u^{-\frac{2}{3}} du$. ## Step 5: Evaluate the integral The integral of $u^{-\frac{2}{3}}$ is $\frac{u^{\frac{1}{3}}}{\frac{1}{3}} = 3u^{\frac{1}{3}}$. So, the integral becomes $\frac{1}{3} \cdot 3u^{\frac{1}{3}} = u^{\frac{1}{3}}$. ## Step 6: Substitute back to the original variable Substituting back $u = 4x - \frac{17}{2}$, we get $(4x - \frac{17}{2})^{\frac{1}{3}}$. ## Step 7: Add a constant to the result Since this is an indefinite integral, we need to add a constant to the result: $(4x - \frac{17}{2})^{\frac{1}{3}} + C$. The final answer is: $\boxed{(4x - \frac{17}{2})^{\frac{1}{3}} + C}$ |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |
| Qwen2-72B-Instruct | To find the indefinite integral of the given function, we first simplify and rewrite it to make integration easier. The function is: \[ f(x) = \frac{4}{3 \sqrt[3]{4x - \frac{17}{2}}^2} \] Let's make a substitution to simplify the expression. Let: \[ u = 4x - \frac{17}{2} \] Then, \[ du = 4dx \quad \Rightarrow \quad dx = \frac{du}{4} \] Substituting \(u\) into the original function gives: \[ f(u) = \frac{4}{3 \sqrt[3]{u}^2} \] This simplifies to: \[ f(u) = \frac{4}{3 u^{2/3}} \] Now, we integrate with respect to \(u\): \[ \int f(u) du = \int \frac{4}{3 u^{2/3}} du \] This is a power rule integral, which can be solved as follows: \[ \int \frac{4}{3 u^{2/3}} du = \frac{4}{3} \int u^{-2/3} du \] Using the power rule for integrals (\(\int x^n dx = \frac{x^{n+1}}{n+1}\) for \(n \neq -1\)): \[ \frac{4}{3} \int u^{-2/3} du = \frac{4}{3} \cdot \frac{u^{1/3}}{1/3} + C \] Simplifying: \[ = 4u^{1/3} + C \] Substitute back \(u = 4x - \frac{17}{2}\): \[ = 4(4x - \frac{17}{2})^{1/3} + C \] Thus, the indefinite integral of the given function is: \[ \boxed{4(4x - \frac{17}{2})^{1/3} + C} \] |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |
| baichuan4 | To find the indefinite integral of the function $\frac{4}{3 \sqrt[3]{4 x-\frac{17}{2}}^2}$, we first need to simplify the expression. Let's start by setting up the integral: $$ \int \frac{4}{3 \sqrt[3]{4 x-\frac{17}{2}}^2} \, dx $$ We can simplify the denominator by recognizing that $\sqrt[3]{4 x-\frac{17}{2}}^2$ is the cube root of $(4 x-\frac{17}{2})^2$. To make the integral easier to handle, we can perform a substitution. Let's define a new variable $u$ such that: $$ u = 4 x - \frac{17}{2} $$ Taking the derivative of $u$ with respect to $x$, we get: $$ du = 4 \, dx $$ Now, we can rewrite the original integral in terms of $u$: $$ \int \frac{4}{3 \sqrt[3]{u}^2} \cdot \frac{1}{4} \, du $$ Simplifying, we have: $$ \int \frac{1}{3 \sqrt[3]{u}^2} \, du $$ This integral can be further simplified by recognizing that $\sqrt[3]{u}^2 = u^{\frac{2}{3}}$. Therefore, the integral becomes: $$ \int \frac{1}{3} u^{-\frac{2}{3}} \, du $$ Now, we can integrate using the power rule for integrals: $$ \int u^{-\frac{2}{3}} \, du = \frac{u^{1 - \frac{2}{3}}}{1 - \frac{2}{3}} + C $$ Simplifying the exponent and the constant, we get: $$ \frac{3}{1} u^{\frac{1}{3}} + C = 3 u^{\frac{1}{3}} + C $$ Finally, we need to substitute back $u$ in terms of $x$: $$ 3 (4 x - \frac{17}{2})^{\frac{1}{3}} + C $$ Therefore, the indefinite integral of the given function is: $$ \boxed{3 (4 x - \frac{17}{2})^{\frac{1}{3}} + C} $$ where $C$ is the constant of integration. |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |
| doubao | Let $u=4x-\frac{17}{2}$, then $du=4dx$. The integral becomes: \[ \begin{align*} &\int\frac{4}{3\sqrt[3]{(4x-\frac{17}{2})^2}}dx\\ =&\frac{1}{3}\int\frac{1}{\sqrt[3]{u^2}}du\\ =&\frac{1}{3}\int u^{-\frac{2}{3}}du\\ =&\frac{1}{3}\cdot\frac{3}{1}u^{\frac{1}{3}}+C\\ =&\sqrt[3]{u}+C\\ =&\sqrt[3]{4x-\frac{17}{2}}+C \end{align*} \] So the answer is $\boxed{\sqrt[3]{4x-\frac{17}{2}}+C}$. |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |
| gpto1 | To find the indefinite integral of the function \(\frac{4}{3 \left( \sqrt[3]{4x - \dfrac{17}{2}} \right)^2}\), we can rewrite the integral in terms of exponents and then apply substitution. First, express the function with exponents: \[ \int \frac{4}{3 \left( \sqrt[3]{4x - \dfrac{17}{2}} \right)^2} \, dx = \frac{4}{3} \int \left(4x - \dfrac{17}{2}\right)^{-2/3} \, dx \] Now, let us make a substitution: Let \( u = 4x - \dfrac{17}{2} \), so \( du = 4 \, dx \) or \( dx = \dfrac{du}{4} \). Substitute into the integral: \[ \frac{4}{3} \int u^{-2/3} \cdot \frac{du}{4} = \frac{1}{3} \int u^{-2/3} \, du \] Now, integrate \( u^{-2/3} \): \[ \int u^{-2/3} \, du = \frac{ u^{1/3} }{1/3} + C = 3u^{1/3} + C \] Multiply by \(\dfrac{1}{3}\): \[ \frac{1}{3} \cdot 3u^{1/3} = u^{1/3} \] Replace \( u = 4x - \dfrac{17}{2} \) and express back in radical form: \[ u^{1/3} = \left( 4x - \dfrac{17}{2} \right)^{1/3} = \sqrt[3]{4x - \dfrac{17}{2}} \] Thus, the indefinite integral is: \[ \int \frac{4}{3 \left( \sqrt[3]{4x - \dfrac{17}{2}} \right)^2} \, dx = \sqrt[3]{4x - \dfrac{17}{2}} + C \] \(\boxed{\,\sqrt[3]{4x\ -\ \dfrac{17}{2}}\ +\ C\,}\) |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |
| kimi | 1. **Identify the function to be integrated:** \[ \int \frac{4}{3 \sqrt[3]{4x - \frac{17}{2}}} \, dx \] 2. **Rewrite the integrand in a more convenient form:** \[ \frac{4}{3 \sqrt[3]{4x - \frac{17}{2}}} = \frac{4}{3} \cdot \frac{1}{\sqrt[3]{4x - \frac{17}{2}}} \] 3. **Substitute \( u = 4x - \frac{17}{2} \):** \[ du = 4 \, dx \quad \Rightarrow \quad dx = \frac{1}{4} \, du \] 4. **Change the limits of integration:** \[ x = 0 \quad \Rightarrow \quad u = 4 \cdot 0 - \frac{17}{2} = -\frac{17}{2} \] \[ x = x \quad \Rightarrow \quad u = 4x - \frac{17}{2} \] 5. **Rewrite the integral in terms of \( u \):** \[ \int \frac{4}{3 \sqrt[3]{4x - \frac{17}{2}}} \, dx = \frac{4}{3} \int \frac{1}{\sqrt[3]{4x - \frac{17}{2}}} \, dx = \frac{4}{3} \int \frac{1}{\sqrt[3]{u}} \cdot \frac{1}{4} \, du \] \[ = \frac{1}{3} \int u^{-\frac{1}{3}} \, du \] 6. **Integrate \( u^{-\frac{1}{3}} \):** \[ \int u^{-\frac{1}{3}} \, du = \frac{u^{\frac{2}{3}}}{\frac{2}{3}} = \frac{3}{2} u^{\frac{2}{3}} \] 7. **Substitute back \( u = 4x - \frac{17}{2} \):** \[ \frac{1}{3} \int u^{-\frac{1}{3}} \, du = \frac{1}{3} \cdot \frac{3}{2} u^{\frac{2}{3}} + C \] \[ = \frac{1}{2} (4x - \frac{17}{2})^{\frac{2}{3}} + C \] 8. **Final answer:** \[ \boxed{\frac{1}{2} (4x - \frac{17}{2})^{\frac{2}{3}} + C} \] |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |
| minimax | To find the indefinite integral of the function $\frac{4}{3 \sqrt[3]{4 x-\frac{17}{2}}^2}$, we can use a substitution method. Let's set: $$ u = 4x - \frac{17}{2} $$ Then, we differentiate both sides with respect to $x$ to find the relationship between $du$ and $dx$: $$ du = 4 \, dx $$ Now, we can rewrite the integral in terms of $u$: $$ \int \frac{4}{3 \sqrt[3]{4 x-\frac{17}{2}}^2} \, dx = \int \frac{4}{3 \sqrt[3]{u}^2} \cdot \frac{1}{4} \, du $$ Simplify the constant factors: $$ \int \frac{1}{3 \sqrt[3]{u}^2} \, du $$ Now, we can integrate with respect to $u$: $$ \int u^{-\frac{2}{3}} \, du = \frac{u^{-\frac{2}{3} + 1}}{-\frac{2}{3} + |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |
| wenxinyiyan | 首先，我们对给定的函数进行简化处理： $\frac{4}{3 \sqrt[3]{4 x-\frac{17}{2}}^2} = \frac{4}{3} \left(4 x-\frac{17}{2}\right)^{-\frac{2}{3}}$ 接下来，我们利用幂函数的积分公式$\int x^n dx = \frac{1}{n+1} x^{n+1} + C$（其中$n \neq -1$）进行积分。 将$-\frac{2}{3}$代入$n$，我们得到： $\int \frac{4}{3} \left(4 x-\frac{17}{2}\right)^{-\frac{2}{3}} dx = \frac{4}{3} \times \frac{1}{-\frac{2}{3}+1} \left(4 x-\frac{17}{2}\right)^{-\frac{2}{3}+1} + C$ 化简后得到： $= 4 \left(4 x-\frac{17}{2}\right)^{\frac{1}{3}} + C$ 所以，该函数的不定积分为： $\boxed{4 \left(4 x-\frac{17}{2}\right)^{\frac{1}{3}} + C}$ |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |
| zhipu | To find the indefinite integral of the function \( \frac{4}{3 \sqrt[3]{4 x - \frac{17}{2}}^2} \), we can follow these steps: 1. **Rewrite the integrand using exponent notation:** The integrand \( \frac{4}{3 \sqrt[3]{4 x - \frac{17}{2}}^2} \) can be rewritten using exponents. Recall that \( \sqrt[3]{y} = y^{1/3} \), so: \[ \sqrt[3]{4 x - \frac{17}{2}}^2 = (4 x - \frac{17}{2})^{2/3} \] Therefore, the integrand becomes: \[ \frac{4}{3 (4 x - \frac{17}{2})^{2/3}} \] 2. **Set up the integral:** The integral we need to solve is: \[ \int \frac{4}{3 (4 x - \frac{17}{2})^{2/3}} \, dx \] 3. **Use a substitution:** Let \( u = 4 x - \frac{17}{2} \). Then, the differential \( du \) is: \[ du = 4 \, dx \quad \Rightarrow \quad dx = \frac{du}{4} \] 4. **Substitute and simplify:** Substitute \( u \) and \( dx \) into the integral: \[ \int \frac{4}{3 u^{2/3}} \cdot \frac{du}{4} = \int \frac{1}{3 u^{2/3}} \, du \] 5. **Integrate using the power rule:** The integral of \( u^{-2/3} \) is: \[ \int u^{-2/3} \, du = \frac{u^{1/3}}{1/3} + C = 3 u^{1/3} + C \] Therefore: \[ \int \frac{1}{3 u^{2/3}} \, du = \frac{1}{3} \cdot 3 u^{1/3} + C = u^{1/3} + C \] 6. **Substitute back \( u \):** Recall \( u = 4 x - \frac{17}{2} \), so: \[ u^{1/3} = (4 x - \frac{17}{2})^{1/3} \] 7. **Write the final answer:** The indefinite integral is: \[ \boxed{(4 x - \frac{17}{2})^{1/3} + C} \] |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |
| deepseek-v3 | To find the indefinite integral of the function: \[ \frac{4}{3 \sqrt[3]{4x - \frac{17}{2}}^2} \] we can follow these steps: ### Step 1: Simplify the Expression First, let's rewrite the integrand in a more manageable form. Notice that: \[ \sqrt[3]{4x - \frac{17}{2}}^2 = \left(4x - \frac{17}{2}\right)^{2/3} \] So, the integrand becomes: \[ \frac{4}{3} \cdot \left(4x - \frac{17}{2}\right)^{-2/3} \] ### Step 2: Use Substitution Let’s use the substitution method. Let: \[ u = 4x - \frac{17}{2} \] Then, the derivative of \( u \) with respect to \( x \) is: \[ \frac{du}{dx} = 4 \quad \Rightarrow \quad du = 4 \, dx \quad \Rightarrow \quad dx = \frac{du}{4} \] ### Step 3: Rewrite the Integral Substituting \( u \) and \( dx \) into the integral: \[ \int \frac{4}{3} \cdot u^{-2/3} \cdot \frac{du}{4} \] The \( 4 \) in the numerator and denominator cancels out: \[ \int \frac{1}{3} \cdot u^{-2/3} \, du \] ### Step 4: Integrate Now, integrate with respect to \( u \): \[ \frac{1}{3} \int u^{-2/3} \, du \] Recall that: \[ \int u^n \, du = \frac{u^{n+1}}{n+1} + C \quad \text{(for } n \neq -1 \text{)} \] Here, \( n = -\frac{2}{3} \), so: \[ \frac{1}{3} \cdot \frac{u^{(-2/3) + 1}}{(-2/3) + 1} + C = \frac{1}{3} \cdot \frac{u^{1/3}}{1/3} + C = u^{1/3} + C \] ### Step 5: Substitute Back Replace \( u \) with the original expression: \[ \left(4x - \frac{17}{2}\right)^{1/3} + C \] ### Step 6: Simplify the Expression The cube root can be written as: \[ \sqrt[3]{4x - \frac{17}{2}} + C \] Thus, the indefinite integral is: \[ \boxed{\sqrt[3]{4x - \frac{17}{2}} + C} \] |
\frac{4 x-\frac{17}{2}}{\sqrt[3]{4 x-\frac{17}{2}}^2} |