The integral ∫ (2x¹² + 5x⁹) / (x⁵ + x³ + 1)³ dx is equal to (a) x⁵ / (2(x⁵ + x³ + 1)²) + C (b) -x¹⁰

Q: Why can't we just use partial fractions?

Partial fractions are designed for rational functions where the denominator can be factored into linear or quadratic factors. Here, the denominator is a high-degree polynomial raised to the third power. Expanding that would lead to a polynomial of degree 15, which is impossible to work with. We need a substitution that simplifies the structure, not expands it.

Q: How did you know to factor out $x^5$ specifically?

It's about the 'degree gap'. The numerator has terms of degree 12 and 9. The denominator has a base of degree 5. When you raise that base to the power of 3, you get degree 15. The difference between the numerator's degree and the denominator's degree is 3. By factoring out $x^5$, we effectively shift the powers to match the derivative of the base.

Q: Why do we get a negative sign in the derivative?

Remember the power rule for differentiation: $\frac{d}{dx}(x^n) = nx^{n-1}$. When $n$ is negative, say $-2$, the derivative is $-2x^{-3}$. The negative sign is a natural consequence of differentiating negative exponents. Always keep an eye on your signs!

Question

The integral $\int \frac{2x^{12} + 5x^9}{(x^5 + x^3 + 1)^3} dx$ is equal to

Options:

(A) $\frac{x^5}{2(x^5 + x^3 + 1)^2} + C$

(B) $\frac{-x^{10}}{2(x^5 + x^3 + 1)^2} + C$

(C) $\frac{-x^5}{(x^5 + x^3 + 1)^2} + C$

(D) $\frac{x^{10}}{2(x^5 + x^3 + 1)^2} + C$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a massive, complex machine. At first glance, it looks like a chaotic mess of gears and levers—in our case, the integral:

$$I = \int \frac{2x^{12} + 5x^9}{(x^5 + x^3 + 1)^3} dx$$

The numerator is a high-degree polynomial, and the denominator is a cubic polynomial raised to the third power. If you try to jump in with a standard substitution, you will quickly find yourself lost in a maze of impossible derivatives.

But here is the secret: every complex integral in JEE Advanced is a puzzle designed to test your ability to see the underlying structure. We do not need to fight the complexity; we need to manipulate it.

### The Art of Algebraic Surgery

When direct substitution fails, we must perform what I call 'algebraic surgery'. We need to force the numerator to become the derivative of the denominator.

We look for the highest power of $x$ inside the denominator's base, which is $x^5$. By factoring $x^5$ out of the expression $(x^5 + x^3 + 1)^3$, we are essentially pulling the 'engine' out of the machine.

We write the denominator as:

$$[x^5(1 + x^{-2} + x^{-5})]^3$$

This simplifies to $x^{15}(1 + x^{-2} + x^{-5})^3$. This $x^{15}$ is the key that unlocks the entire problem, as it is the factor we need to divide the numerator by to reveal the hidden derivative.

### The Transformation

Now, let us look at our integral again:

$$I = \int \frac{2x^{12} + 5x^9}{x^{15}(1 + x^{-2} + x^{-5})^3} dx$$

By distributing the $x^{15}$ into the numerator, we get:

$$\int \frac{2x^{-3} + 5x^{-6}}{(1 + x^{-2} + x^{-5})^3} dx$$

Suddenly, the chaos disappears. We have a function in the denominator and its derivative (almost) in the numerator. This is the moment of clarity we strive for in every JEE problem.

### The Substitution Revelation

Let $t = 1 + x^{-2} + x^{-5}$. When we differentiate this with respect to $x$, we get:

$$\frac{dt}{dx} = -2x^{-3} - 5x^{-6}$$

This means $dt = -(2x^{-3} + 5x^{-6}) dx$. Our numerator is exactly $-(2x^{-3} + 5x^{-6}) dx$, which is $-dt$.

The integral transforms into the elegant form:

$$I = -\int t^{-3} dt$$

This is a standard power rule integration. The result is $-[\frac{t^{-2}}{-2}] + C$, which simplifies to $\frac{1}{2t^2} + C$.

### The Final Polish

We are almost there, but we must return to the world of $x$. Substituting $t = 1 + x^{-2} + x^{-5}$ back into our result, we get:

$$I = \frac{1}{2(1 + x^{-2} + x^{-5})^2} + C$$

To match the options, we simplify the fraction:

$$1 + \frac{1}{x^2} + \frac{1}{x^5} = \frac{x^5 + x^3 + 1}{x^5}$$

Squaring this and taking the reciprocal gives us the final result:

$$\mathbf{I = \frac{x^{10}}{2(x^5 + x^3 + 1)^2} + C}$$

And there it is—the solution, clean and precise. Remember, the complexity of a problem is often just a mask for a simple, elegant truth waiting to be uncovered.

Animated Solution for Mathematics - Indefinite Integration: The integral $\int \frac{2 x ^{12} + 5 x ^{9}}{( x ^{5} + x ^{3} + 1 ) ^{3}} d x$ is equal to

Select Answer:

Visualized Solution

The Given Integral

Strategy: Forcing a Substitution

Factoring $x^{5}$ from the Denominator

Simplifying the Denominator

Adjusting the Numerator

Simplifying the Numerator

Identifying the Substitution $t$

Differentiating $t$

Rewriting the Integral in $t$

Integrating $t^{- 3}$

Back-Substituting $x$

Final Simplification

The Final Answer

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Art of Algebraic Surgery

The Transformation

The Substitution Revelation

The Final Polish

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Animated Solution for Mathematics - Indefinite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The integral ∫(x5+x3+1)32x12+5x9​dx is equal to

Select Answer:

Visualized Solution

The Given Integral

Strategy: Forcing a Substitution

Factoring .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x5 from the Denominator

Simplifying the Denominator

Adjusting the Numerator

Simplifying the Numerator

Identifying the Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

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

Rewriting the Integral in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Integrating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t−3

Back-Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x