Let J_{n, m}=int_{0}^{frac{1}{2}} frac{x^{n}} | vedclass

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(C) Given $J_{n, m}=\int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m}-1} d x$. For $i \leq j$,$a_{i j}=J_{6+i, 3}-J_{i+3,3} = \int_{0}^{\frac{1}{2}} \frac{x^{

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$(105)^{2} 	imes 2^{38}$

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(C) Given $J_{n, m}=\int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m}-1} d x$. For $i \leq j$,$a_{i j}=J_{6+i, 3}-J_{i+3,3} = \int_{0}^{\frac{1}{2}} \frac{x^{6+i}-x^{i+3}}{x^{3}-1} d x = \int_{0}^{\frac{1}{2}} \frac{x^{i+3}(x^{3}-1)}{x^{3}-1} d x = \int_{0}^{\frac{1}{2}} x^{i+3} d x$. Evaluating the integral: $a_{i j} = \left[ \frac{x^{i+4}}{i+4} ight]_{0}^{\frac{1}{2}} = \frac{(1/2)^{i+4}}{i+4}$. Thus,$a_{11} = \frac{(1/2)^{5}}{5} = \frac{1}{5 \cdot 2^{5}}$,$a_{12} = \frac{(1/2)^{5}}{5} = \frac{1}{5 \cdot 2^{5}}$,$a_{13} = \frac{(1/2)^{5}}{5} = \frac{1}{5 \cdot 2^{5}}$. $a_{22} = \frac{(1/2)^{6}}{6} = \frac{1}{6 \cdot 2^{6}}$,$a_{23} = \frac{(1/2)^{6}}{6} = \frac{1}{6 \cdot 2^{6}}$. $a_{33} = \frac{(1/2)^{7}}{7} = \frac{1}{7 \cdot 2^{7}}$. Matrix $A = \begin{bmatrix} \frac{1}{5 \cdot 2^{5}} & \frac{1}{5 \cdot 2^{5}} & \frac{1}{5 \cdot 2^{5}} \ 0 & \frac{1}{6 \cdot 2^{6}} & \frac{1}{6 \cdot 2^{6}} \ 0 & 0 & \frac{1}{7 \cdot 2^{7}} \end{bmatrix}$. $|A| = \frac{1}{5 \cdot 2^{5}} \cdot \frac{1}{6 \cdot 2^{6}} \cdot \frac{1}{7 \cdot 2^{7}} = \frac{1}{210 \cdot 2^{18}}$. We need $|\operatorname{adj} A^{-1}| = |A^{-1}|^{3-1} = |A^{-1}|^{2} = \frac{1}{|A|^{2}} = (210 \cdot 2^{18})^{2} = (2 \cdot 105)^{2} \cdot 2^{36} = 4 \cdot (105)^{2} \cdot 2^{36} = (105)^{2} \cdot 2^{38}$.

Let $J_{n, m}=\int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m}-1} d x, \quad \forall n>m$ and $n, m \in N$. Consider a matrix $A=\left[a_{i j}\right]_{3 \times 3}$ where $a_{i j}=J_{6+i, 3}-J_{i+3,3}$ for $i \leq j$ and $a_{i j}=0$ for $i>j$. Then $\left|\operatorname{adj} A^{-1}\right|$ is:

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