Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{2e^x+3e^{-x}}{4e^x+7e^{-x}}dx=\frac{1}{14}(ux+v{\log }_e(4e^x+7e^{-x}))+C$, where $C$ is a constant of integration, then $u+v$ is equal to.

Mastering Exponential Rational Functions

Welcome, fellow traveler on the JEE Advanced journey. Today, we are going to demystify a problem that often intimidates students at first glance: the integration of a rational function involving exponentials.

Specifically, we are looking at the integral:

At first, it looks like a mess of $e^x$ and $e^{-x}$ terms. But in the world of JEE, complexity is often just a mask for a beautiful, hidden structure. Let us peel back that mask.

The Structural Insight

When you encounter an integral of the form

do not panic. Do not immediately reach for complex substitutions that might lead you into a labyrinth of partial fractions.

Instead, use the 'Numerator Decomposition' technique. We want to force the numerator to reveal its relationship with the denominator.

We define our denominator as $D=4e^x+7e^{-x}$. If we calculate its derivative, $D^{\prime }$, we get $D^{\prime }=4e^x-7e^{-x}$.

Now, we express our numerator $N=2e^x+3e^{-x}$ as a linear combination of $D$ and $D^{\prime }$:

This is the secret key. By doing this, we transform a difficult integral into two simple ones.

The Algebraic Battle

Now, we set up our identity:

To find $A$ and $B$, we compare the coefficients of $e^x$ and $e^{-x}$ on both sides.

For $e^x$, we have $2=4A+4B$, which simplifies to $A+B=\frac{1}{2}$. For $e^{-x}$, we have $3=7A-7B$, which simplifies to $A-B=\frac{3}{7}$.

Solving this system is a breeze. Adding the two equations gives $2A=\frac{1}{2}+\frac{3}{7}=\frac{7+6}{14}=\frac{13}{14}$, so $A=\frac{13}{28}$.

Subtracting them gives $2B=\frac{1}{2}-\frac{3}{7}=\frac{7-6}{14}=\frac{1}{14}$, so $B=\frac{1}{28}$. We have successfully broken down the numerator into manageable pieces.

The Integration

With $A$ and $B$ in hand, our integral becomes:

Substituting our values, we get:

The first part is trivial: $\frac{13}{28}x$. The second part is a classic logarithmic integral, $\int \frac{f^{\prime }(x)}{f(x)}dx={\log }_e|f(x)|+C$.

Thus, the result is:

The Final Match

The problem asks us to match this to the form $\frac{1}{14}(ux+v{\log }_e(4e^x+7e^{-x}))+C$. To do this, we factor out $\frac{1}{14}$ from our result:

Comparing this to the target form, we immediately see that $u=\frac{13}{2}$ and $v=\frac{1}{2}$.

The final step is simply to calculate $u+v=\frac{13}{2}+\frac{1}{2}=\frac{14}{2}=7$.

See? The terror of the exponential integral vanishes when you understand the underlying structure. You didn't just solve a problem; you mastered a technique that will serve you well in many future challenges.

Keep practicing, keep visualizing, and most importantly, keep falling in love with the elegance of mathematics.