The integral ∫ (e^(3 log_e 2x) + 5e^(2 log_e 2x)) / (e^(4 log_e x) + 5e^(3 log_e x) - 7e^(2 log_e x)

Q: Wait, why can't we just integrate the numerator and denominator separately? Like, $\int \frac{A}{B} dx = \frac{\int A dx}{\int B dx}$?

That is a very common trap! Integration is a linear operator, but it does not distribute over division. You cannot integrate the numerator and denominator independently. You must treat the entire expression as a single rational function. Always look for relationships between the numerator and the denominator as a whole.

Q: Why are we allowed to cancel $x^2$ from the numerator and denominator? What if $x=0$?

Great catch! The problem explicitly states $x > 0$. Since $x$ is strictly positive, $x^2$ is never zero, which makes the cancellation mathematically valid. Always check the domain constraints provided in the question; they are there for a reason!

Q: How do I know when to use the $f'(x)/f(x)$ substitution method?

Whenever you see a rational function where the degree of the numerator is related to the derivative of the denominator, always check the derivative of the denominator first. It is the most efficient path to the solution and often saves you from tedious partial fraction decomposition.

Question

The integral $\int \frac{e^{3 \log_e 2x}+5e^{2 \log_e 2x}}{e^{4 \log_e x}+5e^{3 \log_e x}-7e^{2 \log_e x}} dx, x > 0$, is equal to: (where $c$ is a constant of integration)

Options:

(A) $\log_e |x^2 + 5x - 7| + c$

(B) $\frac{1}{2}\log_e |x^2 + 5x - 7| + c$

(C) $4\log_e |x^2 + 5x - 7| + c$

(D) $\log_e \sqrt{x^2 + 5x - 7} + c$

Sigma Percentile · Accepted Answer

### The Logarithmic Veil: Unmasking the Problem

Welcome, fellow traveler on the JEE journey. Today, we face a problem that looks like a monster. When you first glance at the integral

$$\int \frac{e^{3 \log_e 2x}+5e^{2 \log_e 2x}}{e^{4 \log_e x}+5e^{3 \log_e x}-7e^{2 \log_e x}} dx$$

it is natural to feel a surge of anxiety. The exponentials, the logarithms, and the powers create a chaotic appearance, but remember: **complexity is often a disguise.**

In this article, we are going to peel back the layers of this problem, not with brute force, but with the elegance of logarithmic properties. We are going to transform this 'transcendental nightmare' into a simple, beautiful algebraic fraction.

### Phase 1: Stripping the Exponents

To begin, we need our toolkit. Recall the two most powerful identities for this problem:

1. The Power Rule: $n \log_e a = \log_e (a^n)$
2. The Inverse Property: $e^{\log_e f(x)} = f(x)$

Combining these, we get the 'magic wand': $e^{n \log_e f(x)} = (f(x))^n$. Let's apply this to our numerator.

The term $e^{3 \log_e 2x}$ becomes $(2x)^3$, which is $8x^3$. Similarly, $5e^{2 \log_e 2x}$ becomes $5(2x)^2$, which is $5(4x^2) = 20x^2$. Our numerator is now a clean $8x^3 + 20x^2$.

Now, look at the denominator. Using the same logic, $e^{4 \log_e x}$ becomes $x^4$, $5e^{3 \log_e x}$ becomes $5x^3$, and $-7e^{2 \log_e x}$ becomes $-7x^2$.

Suddenly, the 'scary' expression has vanished, leaving us with a standard rational function:

$$I = \int \frac{8x^3 + 20x^2}{x^4 + 5x^3 - 7x^2} dx$$

### Phase 2: The Algebraic Reveal

Now that we have a rational function, we look for common factors. Notice that every term in the numerator and denominator contains at least $x^2$. Since the problem guarantees $x > 0$, we can safely factor out $x^2$ and cancel it:

$$I = \int \frac{x^2(8x + 20)}{x^2(x^2 + 5x - 7)} dx = \int \frac{8x + 20}{x^2 + 5x - 7} dx$$

Take a moment to appreciate this. We have reduced a complex exponential integral into a simple rational integral. This is the heart of JEE problem-solving: simplifying the expression until the path forward becomes obvious.

### Phase 3: The Calculus Insight

Now, we reach the final act. We have the integral $\int \frac{8x + 20}{x^2 + 5x - 7} dx$. Whenever you see a rational function, your first instinct should be to check the derivative of the denominator.

Let $f(x) = x^2 + 5x - 7$. Then, $f'(x) = 2x + 5$.

Look at our numerator: $8x + 20$. If we factor out a $4$, we get $4(2x + 5)$. It is a perfect match!

The numerator is exactly $4$ times the derivative of the denominator. This is the 'Aha!' moment. We can use the substitution $t = x^2 + 5x - 7$, which gives $dt = (2x + 5) dx$.

The integral becomes:

$$I = \int \frac{4}{t} dt = 4 \log_e |t| + c$$

Substituting back $t = x^2 + 5x - 7$, we arrive at our final answer:

**$$4 \log_e |x^2 + 5x - 7| + c$$**

### Final Thoughts

This problem was never about complex integration techniques. It was about pattern recognition and the confidence to simplify.

When you face such problems in the exam, do not panic. Strip away the layers, look for the underlying structure, and trust your tools. You have the power to solve these; just take it one step at a time.

Animated Solution for Mathematics - Indefinite Integration: The integral $\int \frac{e ^{3 l o g_{e} 2 x} + 5 e ^{2 l o g_{e} 2 x}}{e ^{4 l o g_{e} x} + 5 e ^{3 l o g_{e} x} - 7 e ^{2 l o g_{e} x}} d x, x > 0$ , is equal to: (where $c$ is a constant of integration)

Select Answer:

Visualized Solution

Introduction to the Integral

Recalling Logarithmic Properties

Simplifying the Numerator

Simplifying the Denominator

Rewriting the Integral

Factoring out $x^{2}$

Simplifying the Fraction

Checking the Derivative

Applying Substitution

Integrating in terms of $t$

Back Substitution and Final Answer

The Sigma Insight: Integration by Substitution

The Logarithmic Veil

Phase 1

Phase 2

Phase 3

Final Thoughts

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