Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{dx}{(x^2+x+1{)}^2}=a{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+b\left(\frac{2x+1}{x^2+x+1}\right)+C,x>0$ where $C$ is the constant of integration, then the value of $9(\sqrt{3}a+b)$ is equal to.

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are not just solving an integral; we are conducting a symphony.

The problem before us, $I=\int \frac{dx}{(x^2+x+1{)}^2}$, might look like a daunting, impenetrable wall of algebra. But I want you to take a deep breath.

In mathematics, complexity is often just a mask for hidden symmetry. Our job is to peel back that mask.

When you see a quadratic expression like $x^2+x+1$ sitting in the denominator, your intuition should immediately scream, "Complete the square!"

Integration is fundamentally about finding patterns, and a raw quadratic is a chaotic mess. By completing the square, we impose order.

We take the expression $x^2+x+1$ and manipulate it. We add and subtract $\frac{1}{4}$ to create a perfect square trinomial:

Look at that! We have transformed a generic quadratic into a sum of squares. This is the geometric bedrock of our solution.

We are no longer dealing with a random polynomial; we are dealing with a structure that screams for trigonometric intervention.

Now, we enter the most exciting phase of the journey: the substitution. We have the form $u^2+a^2$, where $u=x+\frac{1}{2}$ and $a=\frac{\sqrt{3}}{2}$.

This is the classic signature of the tangent function. Let us set:

This substitution is not arbitrary; it is a key that unlocks the door. If we rearrange this, we get $\tan \theta =\frac{2x+1}{\sqrt{3}}$.

Now, we must transform our differential $dx$. Differentiating both sides with respect to $\theta$, we get:

Imagine the integral now. The denominator becomes ${\left(\frac{3}{4}{\tan }^2\theta +\frac{3}{4}\right)}^2$.

Factoring out the $\frac{3}{4}$, we get $\frac{9}{16}(1+{\tan }^2\theta {)}^2$. Since $1+{\tan }^2\theta ={\sec }^2\theta$, the denominator simplifies to $\frac{9}{16}{\sec }^4\theta$.

The transformation is complete, and the chaos has vanished.

Let us assemble our new integral. Substituting our terms, we have:

Watch the magic happen. The constants $\frac{\sqrt{3}}{2}$ and $\frac{16}{9}$ multiply to give $\frac{8\sqrt{3}}{9}$.

The trigonometric terms $\frac{{\sec }^2\theta }{{\sec }^4\theta }$ simplify to $\frac{1}{{\sec }^2\theta }$, which is simply ${\cos }^2\theta$. Our terrifying integral has collapsed into the elegant form:

To solve this, we use the double-angle identity, ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$. This is a staple of JEE problems—never fear the power of a trigonometric function; just reduce its degree!

Integrating term by term, we get:

We are almost there. We have the answer in terms of $\theta$, but the original question was in terms of $x$. We must return home.

We know $\theta ={\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)$. But what about $\frac{\sin 2\theta }{2}$?

Recall that $\sin 2\theta =2\sin \theta \cos \theta$. Therefore, $\frac{\sin 2\theta }{2}=\sin \theta \cos \theta$.

Using our right-angled triangle where the opposite side is $2x+1$ and the adjacent side is $\sqrt{3}$, the hypotenuse is $2\sqrt{x^2+x+1}$. Thus:

Multiplying these gives us the term $\frac{\sqrt{3}(2x+1)}{4(x^2+x+1)}$. When we distribute the constant $\frac{4\sqrt{3}}{9}$, the math aligns perfectly with the target form provided in the question.

By comparing our result with the target $a{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+b\left(\frac{2x+1}{x^2+x+1}\right)+C$, we identify:

Finally, we calculate the value of $9(\sqrt{3}a+b)$:

And there it is. 15. A beautiful, clean integer at the end of a complex journey.

Remember, in JEE Advanced, the complexity is just a test of your patience and your ability to see the underlying structure. Keep practicing, keep visualizing, and keep falling in love with the process.