∫₀∞ (6 / (e³ˣ + 6e²ˣ + 11eˣ + 6)) dx =
(1) logₑ(32/27)
(2) logₑ(512/81)
(3) logₑ(256/81)
(4) logₑ(30

Q: Wait, why can't we just use substitution $u = e^x$ right at the beginning?

You absolutely can! If you set $u = e^x$, then $du = e^x dx$, which means $dx = \frac{du}{u}$. The integral becomes $\int_1^{\infty} \frac{6}{u(u+1)(u+2)(u+3)} du$. This is also a valid path, but it requires partial fraction decomposition on a quartic denominator. The method shown in the solution is often faster because it avoids the extra $u$ in the denominator.

Q: Why does the upper limit evaluate to zero?

Great question! As $x \to \infty$, the term $e^{-x}$ approaches $0$. When you substitute this into the logarithmic terms like $\ln(1 + a e^{-x})$, you get $\ln(1 + a \cdot 0) = \ln(1)$. Since $\ln(1) = 0$, all the terms at the upper limit vanish. It is a beautiful simplification that happens often in these types of integrals.

Q: How do I know the partial fraction coefficients are correct?

You can use the 'cover-up' method (Heaviside method). For the expression $\frac{6}{(y+1)(y+2)(y+3)}$, to find the coefficient of $\frac{1}{y+1}$, you 'cover up' the $(y+1)$ term and evaluate the rest at $y=-1$. That gives $\frac{6}{(-1+2)(-1+3)} = \frac{6}{(1)(2)} = 3$. It is a very reliable and fast technique!

Question

$\int_0^{\infty} \frac{6}{e^{3x} + 6e^{2x} + 11e^x + 6} dx =$

Options:

(A) $\log_e \left(\frac{32}{27}\right)$

(B) $\log_e \left(\frac{512}{81}\right)$

(C) $\log_e \left(\frac{256}{81}\right)$

(D) $\log_e \left(\frac{302}{27}\right)$

Sigma Percentile · Accepted Answer

### The Beauty of the Exponential Integral

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect an integral that, at first glance, looks like a tangled mess of exponentials.

We are looking at the integral:

$$I = \int_0^{\infty} \frac{6}{e^{3x} + 6e^{2x} + 11e^x + 6} dx$$

It is a classic problem that tests your ability to see structure in complexity.

### Phase 1: The Cubic Denominator

Imagine you are standing before this expression. The denominator $e^{3x} + 6e^{2x} + 11e^x + 6$ might look intimidating, but let us simplify our perspective.

If we let $y = e^x$, the denominator becomes a simple cubic polynomial: $y^3 + 6y^2 + 11y + 6$.

By testing small integer roots or observing the coefficients, we find that the roots are $-1, -2,$ and $-3$. Thus, the polynomial factorizes beautifully into $(y+1)(y+2)(y+3)$.

Returning to our original variable, the integral becomes:

$$I = \int_0^{\infty} \frac{6}{(e^x+1)(e^x+2)(e^x+3)} dx$$

We have already tamed the beast.

### Phase 2: The Art of Decomposition

Now, we face a rational function. We cannot integrate this directly, so we turn to partial fraction decomposition.

We want to express the integrand as a sum of simpler fractions:

$$\frac{6}{(y+1)(y+2)(y+3)} = \frac{A}{y+1} + \frac{B}{y+2} + \frac{C}{y+3}$$

Using the cover-up method, we quickly find $A=3, B=-6,$ and $C=3$. Factoring out the $6$, we get the elegant form:

$$6 \left[ \frac{1}{2(e^x+1)} - \frac{1}{e^x+2} + \frac{1}{2(e^x+3)} ight]$$

This is the turning point. We have broken a complex structure into three manageable pieces.

### Phase 3: The Integration Trick

How do we integrate $\frac{1}{e^x+a}$? The standard trick in our JEE toolkit is to multiply the numerator and denominator by $e^{-x}$.

This gives us $\frac{e^{-x}}{1+ae^{-x}}$. Because the numerator $e^{-x}$ is proportional to the derivative of the denominator $1+ae^{-x}$, we can use the fundamental rule:

$$\int \frac{f'(x)}{f(x)} dx = \ln|f(x)|$$

Applying this to each term, we get:

$$I = 6 \left[ -\frac{1}{2} \ln(1+e^{-x}) + \frac{1}{2} \ln(1+2e^{-x}) - \frac{1}{6} \ln(1+3e^{-x}) ight]_0^{\infty}$$

### Phase 4: The Final Evaluation

Now, we apply the limits. At the upper limit $x 	o \infty$, $e^{-x} 	o 0$, so every term becomes $\ln(1)$, which is $0$.

At the lower limit $x = 0$, $e^{-x} = 1$. Substituting this, we get:

$$I = 0 - 6 \left[ -\frac{1}{2} \ln(2) + \frac{1}{2} \ln(3) - \frac{1}{6} \ln(4) ight]$$

Simplifying the logarithmic expression using the power and quotient rules, we arrive at the final result:

$$\mathbf{I = \ln\left(\frac{32}{27}ight)}$$

### Conclusion

Look at what you have achieved. You took a complex, intimidating integral and, through factorization, decomposition, and a clever algebraic trick, reduced it to a simple logarithmic value.

This is the essence of mathematics—finding the underlying order in apparent chaos. Keep practicing, keep visualizing, and most importantly, keep falling in love with the process.

Animated Solution for Mathematics - Definite Integration: $\int_{0}^{\infty} \frac{6}{e ^{3 x} + 6 e ^{2 x} + 11 e ^{x} + 6} d x =$

Select Answer:

Visualized Solution

Analyze the Integrand

Factorize the Denominator

Partial Fraction Decomposition

Finding Coefficients

Prepare for Integration

Execute Integration

Apply Limits: Upper Bound

Apply Limits: Lower Bound

Simplify Logarithmic Expression

Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Beauty of the Exponential Integral

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫0∞​e3x+6e2x+11ex+66​dx=