Animated Solution for Mathematics - Indefinite Integration: Let $f(x)=\int \frac{7x^{10}+9x^8}{(1+x^2+2x^9{)}^2}dx,x>0,{\lim }_{x\to 0}f(x)=0$ and $f(1)=\frac{1}{4}$. If $A=\left[\begin{array}{ccc}0& 0& 1\\ \frac{1}{4}& f^{\prime }(1)& 1\\ {\alpha }^2& 4& 1\end{array}\right]$ and $B=\text{adj}(\text{adj }A)$ be such that $|B|=81$, then ${\alpha }^2$ is equal to

Welcome, fellow traveler on the path to JEE mastery. Today, we confront a problem that initially looks like a monster.

We have an integral with high-degree polynomials, a matrix with an unknown variable $\alpha$, and a property of adjoints that tests our theoretical depth. But fear not—every complex problem is just a collection of simple, elegant steps waiting to be uncovered.

Let us look at our integral: $f(x)=\int \frac{7x^{10}+9x^8}{(1+x^2+2x^9{)}^2}dx$. When you see high powers of $x$ trapped inside a squared bracket, your intuition should scream: "Factor out the highest power!"

Inside the denominator, the highest power is $x^9$. When we pull $x^9$ out of the bracket, it must be squared, becoming $x^{18}$. To maintain the integrity of our fraction, we divide both the numerator and the denominator by $x^{18}$.

Look at that transformation! The chaos of high powers has settled into a clean, manageable expression with negative exponents. This is the beauty of algebraic manipulation.

Now, let us define our substitution. Let $t=x^{-9}+x^{-7}+2$.

If we differentiate this with respect to $x$, we get $dt=(-9x^{-10}-7x^{-8})dx$. Notice how the numerator is exactly $-(9x^{-10}+7x^{-8})dx$; it is a perfect match!

Substituting back, we get $f(x)=\frac{1}{x^{-9}+x^{-7}+2}+C$. Multiplying the numerator and denominator by $x^9$ gives us $f(x)=\frac{x^9}{1+x^2+2x^9}+C$.

With the condition ${\lim }_{x\to 0}f(x)=0$, we quickly find that $C=0$. Our function is complete: $f(x)=\frac{x^9}{1+x^2+2x^9}$.

To construct matrix $A$, we need $f^{\prime }(1)$. Using the quotient rule on $f(x)$, we find:

Plugging in $x=1$, we get $f^{\prime }(1)=\frac{9(4)-1(20)}{4^2}=\frac{36-20}{16}=1$. Now, our matrix $A$ is defined as:

Expanding the determinant along the first row is a gift. The first two terms are zero, leaving us with $1\cdot (\frac{1}{4}\cdot 4-{\alpha }^2\cdot 1)=1-{\alpha }^2$.

We are given $B=\text{adj}(\text{adj }A)$ and $|B|=81$. The property $|\text{adj}(\text{adj }A)|=|A{|}^{(n-1{)}^2}$ is our final key.

Since $n=3$, we have $|B|=|A{|}^4=81$. This implies $|A|=\pm 3$.

Setting $1-{\alpha }^2=3$ gives ${\alpha }^2=-2$ (impossible for real $\alpha$). Setting $1-{\alpha }^2=-3$ gives ${\alpha }^2=4$.

And there it is. Through calculus and linear algebra, we have arrived at the solution. The final answer is 4.