Let $$f(x)=\int_{}^{} \frac{7x^{10}+9x^{8}}{(1+x^{2}+2x^{9})^{2}}dx, x>0, \lim_{x \rightarrow 0}f(x)=0$$ and $$f(1)=\frac{1}{4.}$$ If $$A= \begin{bmatrix}0 & 0 & 1 \\ \frac{1}{4} & f'(1) & 1 \\ \alpha^{2} & 4 & 1 \end{bmatrix}$$ and B = adj(adj A) be such that |B| = 81 , then $$\alpha^{2}$$ is equal to

 Solve for $$f(x)$$

$$f(x) = \int \frac{7x^{10} + 9x^8}{(1 + x^2 + 2x^9)^2} dx$$. Dividing numerator and denominator by $$x^{18}$$:

$$f(x) = \int \frac{7x^{-8} + 9x^{-10}}{(x^{-9} + x^{-7} + 2)^2} dx$$.

Let $$u = x^{-9} + x^{-7} + 2 \implies du = (-9x^{-10} - 7x^{-8}) dx$$.

$$f(x) = \int \frac{-du}{u^2} = \frac{1}{u} + C = \frac{1}{x^{-9} + x^{-7} + 2} + C = \frac{x^9}{1 + x^2 + 2x^9} + C$$.

Given $$\lim_{x \to 0} f(x) = 0 \implies C = 0$$.

$$f(1) = \frac{1}{1+1+2} = 1/4$$ (Matches).

$$f'(x) = \text{integrand}$$. $$f'(1) = \frac{7+9}{(1+1+2)^2} = \frac{16}{16} = 1$$.

2. Solve Matrix A

$$A = \begin{bmatrix} 0 & 0 & 1 \\ 1/4 & 1 & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix}$$.

$$\det(A) = 1(1 - \alpha^2) = 1 - \alpha^2$$.

For a $$3 \times 3$$ matrix, $$|adj(adj A)| = |A|^{(3-1)^2} = |A|^4$$.

$$|A|^4 = 81 \implies |A| = \pm 3$$.

$$1 - \alpha^2 = 3$$ (No real solution) or $$1 - \alpha^2 = -3 \implies \alpha^2 = 4$$