The value of b 3 for which 12 ∫[3 to b] 1/((x² - 1)(x² - 4)) dx = log_e(49/40), is equal to ______

Q: Why do we use partial fraction decomposition here? Can't we just integrate it directly?

Great question! While you could try other methods, the integrand $\frac{1}{(x^2 - 1)(x^2 - 4)}$ is a classic rational function. Partial fraction decomposition is the most efficient 'surgical' tool to break down complex rational expressions into simpler, standard forms that we already have formulas for. It turns a daunting problem into two straightforward logarithmic integrals.

Q: Why can't we just expand the polynomial in the denominator and solve for $b$?

You absolutely could, but it would lead to a high-degree polynomial equation in $b$, which is often difficult and time-consuming to solve. In competitive exams like JEE, time is your most valuable asset. By keeping the expression in its factored, logarithmic form, we can compare the arguments of the logarithms directly, which is much faster and less prone to algebraic errors.

Q: Why is the integral of $\frac{1}{x^2 - a^2}$ equal to $\frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right|$?

This is a standard result derived from partial fractions! If you decompose $\frac{1}{x^2 - a^2}$ into $\frac{1}{2a} \left( \frac{1}{x - a} - \frac{1}{x + a} \right)$, integrating each term gives $\frac{1}{2a} (\ln |x - a| - \ln |x + a|)$, which simplifies to the formula using the quotient rule for logarithms. It's a fundamental tool in your calculus toolkit—make sure to memorize it!

Question

The value of $b > 3$ for which $12 \int_{3}^{b} \frac{1}{(x^2 - 1)(x^2 - 4)} dx = \log_e \left( \frac{49}{40} \right)$, is equal to ______.

Sigma Percentile · Accepted Answer

### The Beauty of Decomposition: A Calculus Journey

Imagine you are standing at the edge of a mathematical landscape, staring at a complex integral:

$$12 \int_{3}^{b} \frac{1}{(x^2 - 1)(x^2 - 4)} dx = \ln \left( \frac{49}{40} \right)$$

At first glance, it looks intimidating. But in the world of JEE Advanced, intimidation is just an invitation to look closer. Let's break this down together.

### Phase 1: The Art of Partial Fractions

Our integrand, $\frac{1}{(x^2 - 1)(x^2 - 4)}$, is a rational function. The denominator is a product of two differences of squares.

The secret to unlocking this integral lies in partial fraction decomposition. Notice the relationship between the factors: $(x^2 - 1) - (x^2 - 4) = 3$.

This is not a coincidence; it is a gift. We can rewrite the numerator as $\frac{1}{3} [ (x^2 - 1) - (x^2 - 4) ]$. This allows us to split the fraction into two much friendlier terms:

$$\frac{1}{3} \left( \frac{1}{x^2 - 4} - \frac{1}{x^2 - 1} \right)$$

### Phase 2: The Integration Toolkit

Now, we substitute this back into our original equation. We have a constant $12$ outside the integral, and our decomposition gives us a $\frac{1}{3}$. Multiplying them, $12 \cdot \frac{1}{3} = 4$.

Our integral is now:

$$4 \int_{3}^{b} \left( \frac{1}{x^2 - 4} - \frac{1}{x^2 - 1} \right) dx$$

We recall our trusty standard formula: $\int \frac{1}{x^2 - a^2} dx = \frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right|$. Applying this to both terms, we get:

$$4 \left[ \frac{1}{4} \ln \left| \frac{x - 2}{x + 2} \right| - \frac{1}{2} \ln \left| \frac{x - 1}{x + 1} \right| \right]_{3}^{b}$$

### Phase 3: The Logarithmic Dance

Before we plug in the limits, let's simplify. Distributing the $4$ gives us $\ln \left| \frac{x - 2}{x + 2} \right| - 2 \ln \left| \frac{x - 1}{x + 1} \right|$.

To combine these, we use the power rule $n \ln A = \ln A^n$ and the quotient rule $\ln A - \ln B = \ln \frac{A}{B}$. This transforms our expression into a single, elegant logarithm:

$$\ln \left| \frac{(x - 2)(x + 1)^2}{(x + 2)(x - 1)^2} \right|$$

### Phase 4: The Final Reveal

Now, we evaluate this from $3$ to $b$. Plugging in $3$ yields $\ln \left( \frac{4}{5} \right)$.

Setting our expression equal to the right-hand side, $\ln \left( \frac{49}{40} \right)$, we get:

$$\ln \left( \frac{(b - 2)(b + 1)^2}{(b + 2)(b - 1)^2} \right) - \ln \left( \frac{4}{5} \right) = \ln \left( \frac{49}{40} \right)$$

Moving the $\ln \left( \frac{4}{5} \right)$ to the right side and combining, we find the argument must equal $\frac{49}{50}$.

Testing $b = 6$ reveals that:

$$\frac{(6 - 2)(6 + 1)^2}{(6 + 2)(6 - 1)^2} = \frac{4 \cdot 49}{8 \cdot 25} = \frac{49}{50}$$

It matches perfectly! The upper limit **$b = 6$**. Remember, in these problems, elegance often beats brute force. Keep practicing, and you will see these patterns everywhere!

Animated Solution for Mathematics - Definite Integration: The value of $b > 3$ for which $12 \int_{3}^{b} \frac{1}{( x ^{2} - 1 ) ( x ^{2} - 4 )} d x = lo g_{e} (\frac{49}{40})$ , is equal to ______.

Enter Numerical Value:

Visualized Solution

Understanding the Problem

Partial Fraction Decomposition

Simplifying the Constant Multiplier

Standard Integration Formula

Applying the Integration Formula

Distributing the Constant

Combining Logarithmic Terms

Evaluating at the Limits

Setting up the Final Equation

Solving for $b$

Final Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Beauty of Decomposition

Phase 1

Phase 2

Phase 3

Phase 4

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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; } The value of b>3 for which 12∫3b​(x2−1)(x2−4)1​dx=loge​(4049​), is equal to ______.