| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the indefinite integral of the function \(-\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2}\), we will follow a step-by-step approach. First, let's simplify the integrand: \[ -\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2} \] Notice that the denominator can be rewritten using the property of exponents: \[ \sqrt[3]{7-7 x^2}^2 = (7-7 x^2)^{\frac{2}{3}} \] Thus, the integrand becomes: \[ -\frac{14 x}{3 (7-7 x^2)^{\frac{2}{3}}} \] Next, we will use the substitution method to simplify the integral. Let: \[ u = 7 - 7 x^2 \] Then, the differential \(du\) is: \[ du = -14 x \, dx \implies x \, dx = -\frac{du}{14} \] Substituting \(u\) and \(du\) into the integral, we get: \[ -\frac{14 x}{3 (7-7 x^2)^{\frac{2}{3}}} \, dx = -\frac{14 x}{3 u^{\frac{2}{3}}} \, dx \] Replace \(x \, dx\) with \(-\frac{du}{14}\): \[ -\frac{14 x}{3 u^{\frac{2}{3}}} \left(-\frac{du}{14}\right) = \frac{14 x}{3 u^{\frac{2}{3}}} \cdot \frac{du}{14} = \frac{1}{3 u^{\frac{2}{3}}} \, du \] Simplify the integral: \[ \frac{1}{3} \int u^{-\frac{2}{3}} \, du \] Now, integrate \(u^{-\frac{2}{3}}\): \[ \int u^{-\frac{2}{3}} \, du = \frac{u^{-\frac{2}{3} + 1}}{-\frac{2}{3} + 1} = \frac{u^{\frac{1}{3}}}{\frac{1}{3}} = 3 u^{\frac{1}{3}} \] Thus, the integral becomes: \[ \frac{1}{3} \cdot 3 u^{\frac{1}{3}} = u^{\frac{1}{3}} \] Substitute back \(u = 7 - 7 x^2\): \[ (7 - 7 x^2)^{\frac{1}{3}} \] Therefore, the indefinite integral is: \[ \int -\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2} \, dx = (7 - 7 x^2)^{\frac{1}{3}} + C \] Putting the final answer in a box: \[ \boxed{(7 - 7 x^2)^{\frac{1}{3}} + C} \] |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |
| deepseek-r1 | To find the indefinite integral of the function \(-\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2}\), we can use substitution. Let \( u = 7 - 7x^2 \). Then, the derivative of \( u \) with respect to \( x \) is: \[ \frac{du}{dx} = -14x \implies du = -14x \, dx \] Substituting \( u \) and \( du \) into the integral, we get: \[ \int \frac{-14x}{3 (7 - 7x^2)^{2/3}} \, dx = \frac{1}{3} \int u^{-2/3} \, du \] Integrating \( u^{-2/3} \): \[ \frac{1}{3} \int u^{-2/3} \, du = \frac{1}{3} \left( \frac{u^{1/3}}{1/3} \right) + C = u^{1/3} + C \] Substituting back \( u = 7 - 7x^2 \), we get: \[ (7 - 7x^2)^{1/3} + C \] Thus, the indefinite integral is: \[ \boxed{\sqrt[3]{7 - 7x^2} + C} \] |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |
| gpto1 | To find the indefinite integral of the function: \[ -\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2} \] We start by rewriting the integral in terms of powers: \[ \int -\frac{14 x}{3 (7 - 7 x^2)^{2/3}}\, dx \] Factor out constants and simplify the expression: \[ -\frac{14}{3} \int \frac{x}{(7 - 7 x^2)^{2/3}}\, dx \] Pull out the constant \(7^{2/3}\) from the denominator: \[ -\frac{14}{3 \cdot 7^{2/3}} \int \frac{x}{(1 - x^2)^{2/3}}\, dx \] Simplify the constant: \[ -\frac{14}{3 \cdot 7^{2/3}} = -\frac{2 \cdot 7^{1/3}}{3} \] Now, the integral simplifies to: \[ - \frac{2 \cdot 7^{1/3}}{3} \int \frac{x}{(1 - x^2)^{2/3}}\, dx \] Use the substitution \(u = 1 - x^2\), so \(du = -2x\, dx\) and \(x\, dx = -\frac{1}{2}\, du\): \[ \int \frac{x\, dx}{(1 - x^2)^{2/3}} = -\frac{1}{2} \int u^{-2/3}\, du \] Integrate \(u^{-2/3}\): \[ -\frac{1}{2} \int u^{-2/3}\, du = -\frac{3}{2} u^{1/3} + C \] Substitute back \(u = 1 - x^2\): \[ -\frac{3}{2} (1 - x^2)^{1/3} + C \] Multiply back the constant: \[ - \frac{2 \cdot 7^{1/3}}{3} \times \left( - \frac{3}{2} (1 - x^2)^{1/3} \right) = 7^{1/3} (1 - x^2)^{1/3} + C \] Simplify using \( (1 - x^2)^{1/3} = 7^{-1/3} (7 - 7 x^2)^{1/3} \): \[ 7^{1/3} \cdot (1 - x^2)^{1/3} = (7 - 7 x^2)^{1/3} \] Therefore, the indefinite integral is: \[ \boxed{(7 - 7 x^2)^{1/3} + C} \] \(\boxed{(7\ -\ 7x^{2})^{1/3}\ +\ C}\) |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |
| gpt-4o-2024-08-06 | To find the indefinite integral of the function \(-\frac{14x}{3 \sqrt[3]{7-7x^2}^2}\), we start by simplifying the expression and using a suitable substitution. First, rewrite the integrand: \[ -\frac{14x}{3 \sqrt[3]{(7-7x^2)^2}} = -\frac{14x}{3 (7-7x^2)^{2/3}} \] Now, let's use the substitution \( u = 7 - 7x^2 \). Then, the derivative of \( u \) with respect to \( x \) is: \[ \frac{du}{dx} = -14x \quad \Rightarrow \quad du = -14x \, dx \] This substitution implies: \[ dx = \frac{du}{-14x} \] Substituting these into the integral, we have: \[ \int -\frac{14x}{3 (7-7x^2)^{2/3}} \, dx = \int -\frac{14x}{3 u^{2/3}} \cdot \frac{du}{-14x} \] The \(-14x\) terms cancel out, simplifying the integral to: \[ \int \frac{1}{3 u^{2/3}} \, du \] This integral can be rewritten as: \[ \frac{1}{3} \int u^{-2/3} \, du \] The antiderivative of \( u^{-2/3} \) is: \[ \frac{u^{1/3}}{1/3} = 3u^{1/3} \] Thus, the integral becomes: \[ \frac{1}{3} \cdot 3u^{1/3} = u^{1/3} \] Substituting back \( u = 7 - 7x^2 \), we have: \[ (7 - 7x^2)^{1/3} \] Therefore, the indefinite integral is: \[ \boxed{(7 - 7x^2)^{1/3} + C} \] where \( C \) is the constant of integration. |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |
| DeepSeek-Coder-V2-Lite-Instruct | To find the indefinite integral of the function \( -\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2} \), we will follow these steps: 1. **Simplify the integrand**: \[ -\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2} \] Notice that \( 7-7 x^2 = 7(1-x^2) \), so we can rewrite the integrand as: \[ -\frac{14 x}{3 \sqrt[3]{7(1-x^2)}^2} = -\frac{14 x}{3 \cdot 7^{2/3} (1-x^2)^{2/3}} \] Simplifying further: \[ -\frac{14 x}{3 \cdot 7^{2/3} (1-x^2)^{2/3}} = -\frac{14 x}{3 \cdot 7^{2/3} (1-x^2)^{2/3}} = -\frac{14 x}{3 \cdot 7^{2/3} (1-x^2)^{2/3}} \] 2. **Substitution**: Let \( u = 1 - x^2 \). Then \( du = -2x \, dx \) or \( -du = 2x \, dx \). 3. **Rewrite the integral**: \[ -\frac{14 x}{3 \cdot 7^{2/3} (1-x^2)^{2/3}} dx \] becomes \[ -\frac{14 x}{3 \cdot 7^{2/3} u^{2/3}} \cdot \left( -\frac{du}{2x} \right) = \frac{14 x}{3 \cdot 7^{2/3} u^{2/3}} \cdot \frac{du}{2x} \] Simplifying: \[ \frac{14}{3 \cdot 7^{2/3} \cdot 2} \int u^{-2/3} \, du \] 4. **Integrate**: \[ \int u^{-2/3} \, du = \frac{u^{1/3}}{1/3} = 3u^{1/3} \] 5. **Substitute back \( u = 1 - x^2 \)**: \[ 3(1 - x^2)^{1/3} \] 6. **Include the constant of integration**: \[ 3(1 - x^2)^{1/3} + C \] Thus, the indefinite integral is: \[ \boxed{3(1 - x^2)^{1/3} + C} \] |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Identify the integral to be solved We need to find the indefinite integral of the function $-\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2}$. ## Step 2: Choose a suitable substitution Let's use the substitution $u = 7 - 7x^2$. This will simplify the denominator of the function. ## Step 3: Calculate du Calculate $du$ by taking the derivative of $u$ with respect to $x$. We get $du = -14x dx$. ## Step 4: Substitute u and du into the integral Substitute $u$ and $du$ into the integral. We get $\int -\frac{1}{3 \sqrt[3]{u}^2} du$. ## Step 5: Simplify the integral Simplify the integral by rewriting it as $\int -\frac{1}{3} u^{-\frac{2}{3}} du$. ## Step 6: Integrate Integrate the function with respect to $u$. We get $-\frac{1}{3} \cdot \frac{u^{\frac{1}{3}}}{\frac{1}{3}} + C$. ## Step 7: Simplify the result Simplify the result by rewriting it as $-u^{\frac{1}{3}} + C$. ## Step 8: Substitute back u Substitute back $u = 7 - 7x^2$ into the result. We get $-\sqrt[3]{7-7 x^2} + C$. The final answer is: $\boxed{-\sqrt[3]{7-7 x^2} + C}$ |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^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 $-\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2}$. ## Step 2: Rewrite the integral for easier solution Let's rewrite the integral as $\int -\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2} dx$. ## Step 3: Identify a suitable substitution To simplify the integral, we can use the substitution $u = 7 - 7x^2$, which implies $du = -14x dx$. ## Step 4: Apply the substitution With this substitution, the integral becomes $\int \frac{1}{3 \sqrt[3]{u}^2} \cdot -du$. ## Step 5: Simplify the integral This simplifies to $-\frac{1}{3} \int u^{-\frac{2}{3}} du$. ## Step 6: Solve the integral The integral of $u^{-\frac{2}{3}}$ is $\frac{u^{\frac{1}{3}}}{\frac{1}{3}} = 3u^{\frac{1}{3}}$. ## Step 7: Apply the constant multiple rule Multiplying by the constant $-\frac{1}{3}$, we get $-\frac{1}{3} \cdot 3u^{\frac{1}{3}} = -u^{\frac{1}{3}}$. ## Step 8: Substitute back to the original variable Substituting back $u = 7 - 7x^2$, we get $-(7 - 7x^2)^{\frac{1}{3}}$. ## Step 9: Add the constant of integration The final answer should include a constant of integration, so we add $+C$. The final answer is: $\boxed{-(7 - 7x^2)^{\frac{1}{3}} + C}$ |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |
| Qwen2-72B-Instruct | To find the indefinite integral of the given function, we will first simplify and then integrate. Given function: \[f(x) = -\frac{14x}{3\sqrt[3]{(7-7x^2)^2}}\] Let's simplify the denominator: \[\sqrt[3]{(7-7x^2)^2} = \sqrt[3]{49(1-x^2)^2}\] So, the function becomes: \[f(x) = -\frac{14x}{3\sqrt[3]{49(1-x^2)^2}}\] We can further simplify this to: \[f(x) = -\frac{14x}{3\cdot 7^{2/3}(1-x^2)^{2/3}}\] \[f(x) = -\frac{14x}{21(1-x^2)^{2/3}}\] \[f(x) = -\frac{2x}{3(1-x^2)^{2/3}}\] Now, let's integrate this function with respect to \(x\). Let \(u = 1 - x^2\), so \(du = -2x\,dx\). Then, the integral becomes: \[ \int f(x)\,dx = \int -\frac{2x}{3(1-x^2)^{2/3}}\,dx = \int \frac{1}{3u^{2/3}}(-du) \] \[ = \int \frac{1}{3}u^{-2/3}(-du) = -\frac{1}{3}\int u^{-2/3}\,du \] Using the power rule for integration (\(\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\)), we get: \[ -\frac{1}{3}\int u^{-2/3}\,du = -\frac{1}{3}\left(\frac{u^{1/3}}{1/3}\right) + C \] \[ = -\frac{1}{3}\cdot 3u^{1/3} + C = -u^{1/3} + C \] Substituting back \(u = 1 - x^2\): \[ = -(1 - x^2)^{1/3} + C \] Therefore, the indefinite integral of the given function is: \[ \boxed{-\sqrt[3]{1 - x^2} + C} \] |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |
| baichuan4 | To find the indefinite integral of the function $-\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2}$, we can use a substitution to simplify the integrand. First, let's denote the cubic root of $7-7x^2$ as $u$, so we have: $$ u = \sqrt[3]{7-7x^2} $$ Taking the derivative of $u$ with respect to $x$, we get: $$ \frac{du}{dx} = \frac{d}{dx} \sqrt[3]{7-7x^2} = \frac{-14x}{3\sqrt[3]{(7-7x^2)^2}} $$ Now, we can rewrite the original integrand in terms of $u$ and $du$: $$ -\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2} = -\frac{14 x}{3 u^2} = \frac{du}{dx} $$ Therefore, the integral becomes: $$ \int -\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2} dx = \int \frac{du}{u^2} $$ This is a standard integral that we can solve directly: $$ \int \frac{du}{u^2} = \int u^{-2} du = -u^{-1} + C $$ Where $C$ is the constant of integration. Now, we need to express the solution back in terms of $x$. Using our substitution, we have: $$ -u^{-1} + C = -\frac{1}{\sqrt[3]{7-7x^2}} + C $$ So, the indefinite integral of the given function is: $$ \boxed{-\frac{1}{\sqrt[3]{7-7x^2}} + C} $$ |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |
| doubao | Let $u=7-7x^2$, then $du=-14x dx$. So the integral becomes: \[ \begin{align*} &\int-\frac{14x}{3\sqrt[3]{7-7x^2}^2}dx\\ =&\int-\frac{1}{3u^{\frac{2}{3}}}du\\ =&-\frac{1}{3}\int u^{-\frac{2}{3}}du\\ =&-\frac{1}{3}\cdot\frac{3}{1}u^{\frac{1}{3}}+C\\ =&-\sqrt[3]{u}+C\\ =&-\sqrt[3]{7-7x^2}+C \end{align*} \] Putting it in the box: $\boxed{-\sqrt[3]{7-7x^2}+C}$. |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |
| gpto1 | To find the indefinite integral of the function: \[ -\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2} \] We start by rewriting the integral in terms of powers: \[ \int -\frac{14 x}{3 (7 - 7 x^2)^{2/3}}\, dx \] Factor out constants and simplify the expression: \[ -\frac{14}{3} \int \frac{x}{(7 - 7 x^2)^{2/3}}\, dx \] Pull out the constant \(7^{2/3}\) from the denominator: \[ -\frac{14}{3 \cdot 7^{2/3}} \int \frac{x}{(1 - x^2)^{2/3}}\, dx \] Simplify the constant: \[ -\frac{14}{3 \cdot 7^{2/3}} = -\frac{2 \cdot 7^{1/3}}{3} \] Now, the integral simplifies to: \[ - \frac{2 \cdot 7^{1/3}}{3} \int \frac{x}{(1 - x^2)^{2/3}}\, dx \] Use the substitution \(u = 1 - x^2\), so \(du = -2x\, dx\) and \(x\, dx = -\frac{1}{2}\, du\): \[ \int \frac{x\, dx}{(1 - x^2)^{2/3}} = -\frac{1}{2} \int u^{-2/3}\, du \] Integrate \(u^{-2/3}\): \[ -\frac{1}{2} \int u^{-2/3}\, du = -\frac{3}{2} u^{1/3} + C \] Substitute back \(u = 1 - x^2\): \[ -\frac{3}{2} (1 - x^2)^{1/3} + C \] Multiply back the constant: \[ - \frac{2 \cdot 7^{1/3}}{3} \times \left( - \frac{3}{2} (1 - x^2)^{1/3} \right) = 7^{1/3} (1 - x^2)^{1/3} + C \] Simplify using \( (1 - x^2)^{1/3} = 7^{-1/3} (7 - 7 x^2)^{1/3} \): \[ 7^{1/3} \cdot (1 - x^2)^{1/3} = (7 - 7 x^2)^{1/3} \] Therefore, the indefinite integral is: \[ \boxed{(7 - 7 x^2)^{1/3} + C} \] \(\boxed{(7\ -\ 7x^{2})^{1/3}\ +\ C}\) |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |
| kimi | 1. **Identify the function to integrate:** \[ \int -\frac{14x}{3 \sqrt[3]{7-7x^2}^2} \, dx \] 2. **Simplify the integrand:** \[ -\frac{14x}{3 \sqrt[3]{7-7x^2}^2} = -\frac{14x}{3 \cdot (7-7x^2)^{\frac{2}{3}}} \] 3. **Extract constants and simplify the exponent:** \[ -\frac{14x}{3 (7-7x^2)^{\frac{2}{3}}} = -\frac{14}{3} \cdot \frac{x}{(7-7x^2)^{\frac{2}{3}}} \] 4. **Rewrite the integrand using substitution:** Let \( u = 7 - 7x^2 \). Then \( du = -14x \, dx \). 5. **Solve for \( dx \) in terms of \( du \):** \[ dx = -\frac{1}{14x} \, du \] 6. **Substitute \( u \) and \( dx \) into the integral:** \[ \int -\frac{14}{3} \cdot \frac{x}{(7-7x^2)^{\frac{2}{3}}} \, dx = -\frac{14}{3} \int \frac{x}{u^{\frac{2}{3}}} \cdot \left(-\frac{1}{14x}\right) \, du \] 7. **Simplify the integrand:** \[ -\frac{14}{3} \cdot \left(-\frac{1}{14}\right) \int u^{-\frac{2}{3}} \, du = \frac{1}{3} \int u^{-\frac{2}{3}} \, du \] 8. **Integrate \( u^{-\frac{2}{3}} \):** \[ \int u^{-\frac{2}{3}} \, du = \frac{u^{\frac{1}{3}}}{\frac{1}{3}} = 3u^{\frac{1}{3}} \] 9. **Substitute back \( u = 7 - 7x^2 \):** \[ 3u^{\frac{1}{3}} = 3(7 - 7x^2)^{\frac{1}{3}} \] 10. **Add the constant of integration \( C \):** \[ \frac{1}{3} \cdot 3(7 - 7x^2)^{\frac{1}{3}} + C = (7 - 7x^2)^{\frac{1}{3}} + C \] ### Conclusion: \[ \boxed{(7-7x^2)^{\frac{1}{3}} + C} \] |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |
| minimax | To find the indefinite integral of the function $-\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2}$, we can use a substitution method. Let's set: $$ u = 7 - 7x^2 $$ Then, we differentiate both sides with respect to $x$ to find $du$: $$ du = -14x \, dx $$ Now we can rewrite the integral in terms of $u$: $$ \int -\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2} \, dx = \int -\frac{1}{3} u^{-2/3} \, du $$ We can now integrate with respect to $u$: $$ \int -\frac{1}{3} u^{-2/3} \, du = -\frac{1}{3} \int u^{-2/3} \, du = -\frac{1}{3} \cdot \frac{u^{-2/3 + 1}}{-2/3 + 1} + C $$ Simplify the exponent: $$ = -\frac{1}{3} \cdot \ |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |
| wenxinyiyan | 首先，我们观察给定的函数$-\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2}$，可以尝试通过换元法来简化这个函数。 令$u = 7 - 7x^2$，则$du = -14x \, dx$。 将$u$代入原函数，得到： $-\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2} = -\frac{1}{3} \times \frac{-14x}{\sqrt[3]{7-7x^2}^2} = -\frac{1}{3} \times \frac{1}{\sqrt[3]{u}^2} \times du$ 这样，原函数就被简化为$-\frac{1}{3} \times \frac{1}{\sqrt[3]{u}^2} \times du$。 接下来，我们对其进行积分： $\int -\frac{1}{3} \times \frac{1}{\sqrt[3]{u}^2} \times du = -\frac{1}{3} \int u^{-\frac{2}{3}} \, du$ 利用积分公式$\int x^n \, dx = \frac{1}{n+1} x^{n+1} + C$，得到： $-\frac{1}{3} \int u^{-\frac{2}{3}} \, du = -\frac{1}{3} \times \frac{3}{1} u^{\frac{1}{3}} + C = -\sqrt[3]{u} + C$ 最后，将$u = 7 - 7x^2$代回，得到： $-\sqrt[3]{7 - 7x^2} + C$ 所以，$-\frac{14 x}{3 \sqrt[3]{7-7 x^2}^2}$的一个不定积分为$-\sqrt[3]{7 - 7x^2} + C$。 |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |
| zhipu | To find the indefinite integral of the function \( -\frac{14 x}{3 \sqrt[3]{(7-7 x^2)^2}} \), we can proceed as follows: 1. **Simplify the integrand:** \[ -\frac{14 x}{3 \sqrt[3]{(7-7 x^2)^2}} = -\frac{14 x}{3 (7-7 x^2)^{\frac{2}{3}}} \] 2. **Factor out the constant \( 7 \) from the denominator:** \[ -\frac{14 x}{3 (7 (1 - x^2))^{\frac{2}{3}}} = -\frac{14 x}{3 \cdot 7^{\frac{2}{3}} (1 - x^2)^{\frac{2}{3}}} = -\frac{14 x}{3 \cdot 7^{\frac{2}{3}}} (1 - x^2)^{-\frac{2}{3}} \] 3. **Rewrite the constant factor:** \[ -\frac{14 x}{3 \cdot 7^{\frac{2}{3}}} = -\frac{14}{3 \cdot 7^{\frac{2}{3}}} x \] Let \( C = \frac{14}{3 \cdot 7^{\frac{2}{3}}} \), so the integrand becomes: \[ -C x (1 - x^2)^{-\frac{2}{3}} \] 4. **Use substitution to integrate:** Let \( u = 1 - x^2 \). Then, \( du = -2x \, dx \) or \( dx = -\frac{du}{2x} \). Substituting \( u \) and \( dx \) into the integral: \[ \int -C x (1 - x^2)^{-\frac{2}{3}} \, dx = \int -C x u^{-\frac{2}{3}} \left(-\frac{du}{2x}\right) \] 5. **Simplify the integral:** \[ \int -C x u^{-\frac{2}{3}} \left(-\frac{du}{2x}\right) = \int \frac{C}{2} u^{-\frac{2}{3}} \, du \] 6. **Integrate with respect to \( u \):** \[ \int \frac{C}{2} u^{-\frac{2}{3}} \, du = \frac{C}{2} \int u^{-\frac{2}{3}} \, du \] The antiderivative of \( u^{-\frac{2}{3}} \) is \( \frac{u^{\frac{1}{3}}}{\frac{1}{3}} = 3 u^{\frac{1}{3}} \). Therefore: \[ \frac{C}{2} \cdot 3 u^{\frac{1}{3}} = \frac{3C}{2} u^{\frac{1}{3}} \] 7. **Substitute back \( u = 1 - x^2 \):** \[ \frac{3C}{2} (1 - x^2)^{\frac{1}{3}} \] 8. **Include the constant of integration \( C_1 \):** \[ \frac{3C}{2} (1 - x^2)^{\frac{1}{3}} + C_1 \] 9. **Rewrite the constant \( C \):** Recall \( C = \frac{14}{3 \cdot 7^{\frac{2}{3}}} \), so: \[ \frac{3 \cdot \frac{14}{3 \cdot 7^{\frac{2}{3}}}}{2} = \frac{14}{2 \cdot 7^{\frac{2}{3}}} = \frac{7}{7^{\frac{2}{3}}} \] 10. **Simplify the final expression:** \[ \frac{7}{7^{\frac{2}{3}}} (1 - x^2)^{\frac{1}{3}} + C_1 = 7^{1 - \frac{2}{3}} (1 - x^2)^{\frac{1}{3}} + C_1 = 7^{\frac{1}{3}} (1 - x^2)^{\frac{1}{3}} + C_1 \] Thus, the indefinite integral is: \[ \boxed{7^{\frac{1}{3}} (1 - x^2)^{\frac{1}{3}} + C_1} \] |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |
| deepseek-v3 | To find the indefinite integral of the function: \[ -\frac{14x}{3 \sqrt[3]{7 - 7x^2}^2} \] we can proceed with the following steps. ### Step 1: Simplify the Function First, let's simplify the expression: \[ \sqrt[3]{7 - 7x^2}^2 = (7 - 7x^2)^{2/3} \] So, the function becomes: \[ -\frac{14x}{3 (7 - 7x^2)^{2/3}} \] ### Step 2: Substitute to Simplify the Integral Let’s use substitution to make the integral more manageable. Let: \[ u = 7 - 7x^2 \] Then, the derivative of \( u \) with respect to \( x \) is: \[ \frac{du}{dx} = -14x \quad \Rightarrow \quad du = -14x \, dx \] Notice that the numerator of the integrand, \( -14x \, dx \), is exactly \( du \). Therefore, the integral simplifies to: \[ \int -\frac{14x}{3 u^{2/3}} \, dx = \int \frac{1}{3 u^{2/3}} \, du \] ### Step 3: Integrate with Respect to \( u \) Now, integrate \( \frac{1}{3 u^{2/3}} \) with respect to \( u \): \[ \int \frac{1}{3 u^{2/3}} \, du = \frac{1}{3} \int u^{-2/3} \, du \] The integral of \( u^{-2/3} \) is: \[ \int u^{-2/3} \, du = 3u^{1/3} + C \] So, multiplying by \( \frac{1}{3} \): \[ \frac{1}{3} \cdot 3u^{1/3} + C = u^{1/3} + C \] ### Step 4: Substitute Back for \( u \) Replace \( u \) with \( 7 - 7x^2 \): \[ u^{1/3} + C = \sqrt[3]{7 - 7x^2} + C \] ### Final Answer Thus, the indefinite integral is: \[ \boxed{\sqrt[3]{7 - 7x^2} + C} \] where \( C \) is the constant of integration. |
\frac{7-7 x^2}{\sqrt[3]{7-7 x^2}^2} |