For real numbers α, β, γ and δ, if ∫ ((x² - 1) + tan⁻¹((x²+1)/x)) / ((x⁴ + 3x² + 1)tan⁻¹((x²+1)/x))

Q: Wait, why do we split the integral? Isn't that making it more complicated?

It is a common misconception that splitting an integral makes it harder. In this case, the denominator is a product of two functions. By splitting it, we isolate the terms such that the inverse trigonometric function in the numerator cancels out the one in the denominator, turning a 'monster' integral into two much more standard, solvable forms. It is the classic 'divide and conquer' strategy.

Q: How do I know when to divide the numerator and denominator by $x^2$?

Whenever you encounter a rational function where the denominator is of the form $x^4 + ax^2 + b$ (a bi-quadratic form), dividing by $x^2$ is almost always the correct path. It transforms the expression into one involving $x + \frac{1}{x}$ or $x - \frac{1}{x}$, which are much easier to handle with substitution.

Q: Is the constant $C$ important for finding the values of $\alpha, \beta, \gamma, \delta$?

The constant $C$ is the constant of integration and is essential for the indefinite integral, but it does not affect the coefficients $\alpha, \beta, \gamma, \delta$. You can safely ignore it when comparing your result to the given form to find those specific values.

Question

For real numbers $\alpha, \beta, \gamma$ and $\delta$, if $\int \frac{(x^2 - 1) + 	an^{-1}\left(\frac{x^2+1}{x}ight)}{(x^4 + 3x^2 + 1)	an^{-1}\left(\frac{x^2+1}{x}ight)} dx = \alpha \log_e\left(	an^{-1}\left(\frac{x^2+1}{x}ight)ight) + \beta 	an^{-1}\left(\frac{\gamma(x^2-1)}{x}ight) + \delta 	an^{-1}\left(\frac{x^2+1}{x}ight) + C$ where $C$ is an arbitrary constant, then the value of $10(\alpha + \beta\gamma + \delta)$ is equal to ____

Sigma Percentile · Accepted Answer

### The Art of Decomposition: Taming the Monster Integral

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to confront a problem that, at first glance, looks like a nightmare.

You see an integral like:

$$I = \int \frac{(x^2 - 1) + 	an^{-1}\left(\frac{x^2+1}{x}ight)}{(x^4 + 3x^2 + 1)	an^{-1}\left(\frac{x^2+1}{x}ight)} dx$$

Your instinct might be to panic. But I want you to take a deep breath. In the world of JEE Advanced, the most intimidating problems are often just simple problems wearing a scary mask. Our job is to unmask them.

### Phase 1: The Strategy of Splitting

Look closely at the numerator. It is not a single, monolithic block; it is composed of two distinct parts: $(x^2 - 1)$ and $	an^{-1}\left(\frac{x^2+1}{x}ight)$.

When you see a complex fraction where the numerator is a sum, the most powerful tool in your arsenal is to split the integral. Let us define $I = I_1 + I_2$, where $I_1$ takes the first part of the numerator and $I_2$ takes the second.

Notice the magic in $I_2$: the $	an^{-1}\left(\frac{x^2+1}{x}ight)$ term in the numerator perfectly cancels the one in the denominator. Suddenly, $I_2$ is just:

$$I_2 = \int \frac{1}{x^4+3x^2+1} dx$$

We have already simplified the problem by half!

### Phase 2: Taming the First Beast ($I_1$)

Now, let us focus on $I_1$:

$$I_1 = \int \frac{x^2-1}{(x^4+3x^2+1)	an^{-1}\left(x+\frac{1}{x}ight)} dx$$

This still looks messy. Here is a classic JEE trick: when you see $x^4$ in the denominator, divide both the numerator and the denominator by $x^2$.

This transforms the numerator into $1 - \frac{1}{x^2}$ and the denominator into $x^2(x^2+3+\frac{1}{x^2})	an^{-1}(x+\frac{1}{x})$. By using the identity $x^2 + \frac{1}{x^2} = (x+\frac{1}{x})^2 - 2$, the denominator becomes:

$$((x+\frac{1}{x})^2 + 1)	an^{-1}\left(x+\frac{1}{x}ight)$$

Now, if we set $t = 	an^{-1}(x+\frac{1}{x})$, its derivative $dt$ is exactly $\frac{1}{1+(x+\frac{1}{x})^2} \cdot (1-\frac{1}{x^2}) dx$. The entire integral $I_1$ collapses into $\int \frac{dt}{t}$, which is simply $\ln |t| + C_1$. We have tamed the first beast!

### Phase 3: The Bi-quadratic Challenge ($I_2$)

Now for $I_2 = \int \frac{1}{x^4+3x^2+1} dx$. This is a classic bi-quadratic form. We use the brilliant trick of writing the numerator as $1 = \frac{1}{2}((x^2+1) - (x^2-1))$.

This allows us to split $I_2$ into two integrals, $I_{2a}$ and $I_{2b}$. For $I_{2a}$, we divide by $x^2$ to get:

$$\frac{1}{2} \int \frac{1+1/x^2}{x^2+3+1/x^2} dx = \frac{1}{2} \int \frac{1+1/x^2}{(x-1/x)^2+5} dx$$

This is a standard $	an^{-1}$ form. Similarly, $I_{2b}$ becomes:

$$-\frac{1}{2} \int \frac{1-1/x^2}{(x+1/x)^2+1} dx$$

This is also a standard $	an^{-1}$ form. By solving these, we get our final components.

### Phase 4: The Final Synthesis

We have $I_1 = \ln |	an^{-1}(x+\frac{1}{x})|$, $I_{2a} = \frac{1}{2\sqrt{5}} 	an^{-1}(\frac{x^2-1}{\sqrt{5}x})$, and $I_{2b} = -\frac{1}{2} 	an^{-1}(\frac{x^2+1}{x})$.

Comparing this to the given form, we identify $\alpha = 1$, $\beta = \frac{1}{2\sqrt{5}}$, $\gamma = \frac{1}{\sqrt{5}}$, and $\delta = -\frac{1}{2}$. The final calculation is:

$$10(\alpha + \beta\gamma + \delta) = 10\left(1 + \frac{1}{10} - \frac{1}{2}ight) = 6$$

You have done it! You have navigated the complexity and emerged victorious. Keep this spirit of decomposition alive, and no integral will ever be too large for you to handle. **The final answer is 6.**

Enter Numerical Value:

Visualized Solution

Analyzing the Integrand

Splitting the Integral

Simplifying the First Part $I_{1}$

Perfecting the Denominator of $I_{1}$

Substitution for $I_{1}$

Integrating $I_{1}$

Handling the Second Part $I_{2}$

Splitting $I_{2}$ into $I_{2 a}$ and $I_{2 b}$

Solving $I_{2 a}$

Solving $I_{2 b}$

Combining All Results

Comparing with Given Form

Final Calculation

Summary and Takeaway

The Sigma Insight: Evaluation of Special Integral Forms

The Art of Decomposition

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

Analyzing the Integrand

Splitting the Integral

Simplifying the First Part .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Perfecting the Denominator of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Substitution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Integrating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​

Handling the Second Part .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Solving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2a​

Solving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2b​