Animated Solution for Mathematics - Indefinite Integration: If $f(x)=\int \frac{5x^8+7x^6}{(x^2+1+2x^7{)}^2}dx,(x\ge 0),f(0)=0$ and $f(1)=\frac{1}{K}$, then the value of $K$ is

Have you ever looked at an integral and felt like it was staring back at you, daring you to try and solve it? That is exactly how this problem feels. We are faced with $f(x)=\int \frac{5x^8+7x^6}{(x^2+1+2x^7{)}^2}dx$.

It looks like a nightmare of powers and brackets. But in JEE Advanced, complexity is often just a mask for elegance. Let's peel back that mask together.

Before we jump into any calculations, let's look at the structure. We have a rational function with a squared denominator. In calculus, when you see a complex expression raised to a power in the denominator, your first instinct should be: "Can I make the numerator look like the derivative of the inside of that bracket?"

If we let $u=x^2+1+2x^7$, then $du=(2x+14x^6)dx$. That does not match our numerator ($5x^8+7x^6$). This tells us that a direct substitution will not work, and we need to perform some "algebraic surgery" first.

This is where the magic happens. We look at the expression inside the bracket: $x^2+1+2x^7$. The highest power here is $x^7$. Let's force it out.

Because the whole expression is squared, the $x^7$ term comes out as $(x^7{)}^2$, which is $x^{14}$. Now our denominator looks like this:

Now, we take that $x^{14}$ and move it to the numerator. We divide the original numerator by $x^{14}$:

Suddenly, the integral looks completely different:

Now, let's look at the expression inside the bracket again: $t=x^{-5}+x^{-7}+2$. If we differentiate this with respect to $x$, we get:

Look at that! It is almost exactly our numerator. If we multiply by $-1$, we get $-(5x^{-6}+7x^{-8})=\frac{dt}{dx}$. This means our numerator times $dx$ is just $-dt$.

The entire, terrifying integral has collapsed into:

Integrating $t^{-2}$ is a standard power rule application. The integral of $t^n$ is $\frac{t^{n+1}}{n+1}$. Here, $n=-2$, so we get:

Now, we back-substitute $t=x^{-5}+x^{-7}+2$:

To make this look cleaner, multiply the numerator and denominator by $x^7$:

We are given $f(0)=0$. Plugging in $x=0$ gives us $0+C=0$, so $C=0$. Finally, we evaluate $f(1)$:

Since $f(1)=\frac{1}{K}$, we conclude that $K=4$.