Question

If $\int \frac{dx}{x^3(x^{12}+1)^{\frac{5}{6}}} = k_1 \frac{(1+x^{12})^{\frac{1}{6}}}{x^{k_2}}$, where $k_1, k_2 \in R$ then the product $k_1 k_2 =$

• A. -2
• B. -1
• C. 3
• D. -12

Answer 1

Answer: (B)

-1

Answer 2

To evaluate the integral $\int \frac{dx}{x^3(x^{12}+1)^{\frac{5}{6}}}$, we follow these steps:

1. Rewrite the integrand: Factor out $x^{12}$ from the term $(x^{12}+1)^{\frac{5}{6}}$ in the denominator. $(x^{12}+1)^{\frac{5}{6}} = \left(x^{12}\left(1+\frac{1}{x^{12}}\right)\right)^{\frac{5}{6}} = (x^{12})^{\frac{5}{6}} (1+x^{-12})^{\frac{5}{6}} = x^{10} (1+x^{-12})^{\frac{5}{6}}$ Now substitute this back into the integral: $\int \frac{dx}{x^3 \cdot x^{10} (1+x^{-12})^{\frac{5}{6}}} = \int \frac{dx}{x^{13} (1+x^{-12})^{\frac{5}{6}}}$ This can be written as: $\int x^{-13} (1+x^{-12})^{-\frac{5}{6}} dx$

2. Perform substitution: Let $u = 1+x^{-12}$. Differentiate $u$ with respect to $x$: $\frac{du}{dx} = -12x^{-13}$ So, $du = -12x^{-13} dx$, which implies $x^{-13} dx = -\frac{1}{12} du$.

3. Integrate with respect to $u$: Substitute $u$ and $du$ into the integral: $\int u^{-\frac{5}{6}} \left(-\frac{1}{12}\right) du = -\frac{1}{12} \int u^{-\frac{5}{6}} du$ Now integrate $u^{-\frac{5}{6}}$: $\int u^{-\frac{5}{6}} du = \frac{u^{-\frac{5}{6}+1}}{-\frac{5}{6}+1} + C = \frac{u^{\frac{1}{6}}}{\frac{1}{6}} + C = 6u^{\frac{1}{6}} + C$ Substitute this back into the expression for the integral: $-\frac{1}{12} (6u^{\frac{1}{6}}) + C = -\frac{1}{2} u^{\frac{1}{6}} + C$

4. Substitute back to $x$: Replace $u$ with $1+x^{-12}$: $-\frac{1}{2} (1+x^{-12})^{\frac{1}{6}} + C$ Rewrite $(1+x^{-12})^{\frac{1}{6}}$ to match the given form: $(1+x^{-12})^{\frac{1}{6}} = \left(\frac{x^{12}+1}{x^{12}}\right)^{\frac{1}{6}} = \frac{(x^{12}+1)^{\frac{1}{6}}}{(x^{12})^{\frac{1}{6}}} = \frac{(x^{12}+1)^{\frac{1}{6}}}{x^2}$ So the integral becomes: $\int \frac{dx}{x^3(x^{12}+1)^{\frac{5}{6}}} = -\frac{1}{2} \frac{(x^{12}+1)^{\frac{1}{6}}}{x^2} + C$

5. Identify $k_1$ and $k_2$: The problem states that the integral is equal to $k_1 \frac{(1+x^{12})^{\frac{1}{6}}}{x^{k_2}}$. Comparing our result with the given form: $k_1 = -\frac{1}{2}$ $k_2 = 2$

6. Calculate the product $k_1 k_2$: $k_1 k_2 = \left(-\frac{1}{2}\right) \times 2 = -1$

The final answer is $\boxed{-1}$.

Explanation of the solution: The integral $\int \frac{dx}{x^3(x^{12}+1)^{\frac{5}{6}}}$ is solved by first rewriting the denominator as $x^{13}(1+x^{-12})^{\frac{5}{6}}$. Then, a substitution $u = 1+x^{-12}$ is made, which transforms $x^{-13}dx$ into $-\frac{1}{12}du$. The integral becomes $-\frac{1}{12}\int u^{-\frac{5}{6}}du$, which evaluates to $-\frac{1}{2}u^{\frac{1}{6}}$. Substituting back $u = 1+x^{-12}$ and simplifying, we get $-\frac{1}{2}\frac{(x^{12}+1)^{\frac{1}{6}}}{x^2}$. Comparing this with the given form $k_1 \frac{(1+x^{12})^{\frac{1}{6}}}{x^{k_2}}$, we find $k_1 = -\frac{1}{2}$ and $k_2 = 2$. The product $k_1 k_2 = (-\frac{1}{2})(2) = -1$.