Animated Solution for Mathematics - Indefinite Integration: If $\int (e^{2x}+2e^x-e^{-x}-1)e^{(e^x+e^{-x})}dx=g(x)e^{(e^x+e^{-x})}+c$, where $c$ is a constant of integration, then $g(0)$ is equal to :

Taming the Exponential Beast

Imagine you are sitting in the examination hall. You turn the page, and there it is: a daunting integral involving $e^{2x}$, $e^x$, and $e^{-x}$ all tangled together. Your first instinct might be panic.

You might think, "Do I need to use Integration by Parts? Do I need partial fractions?" Take a deep breath. In JEE Advanced, the most intimidating problems are often the ones that hide the most elegant solutions.

This problem is not a test of your endurance; it is a test of your vision.

Simplifying the Landscape

When you see expressions like $e^{2x}$, $e^x$, and $e^{-x}$ appearing repeatedly, your brain should immediately scream: Substitution! Let us simplify the landscape by setting $e^x=t$.

This is a classic maneuver. If $e^x=t$, then $dx=\frac{dt}{t}$.

Let us watch the transformation. The integral $I=\int (e^{2x}+2e^x-e^{-x}-1)e^{(e^x+e^{-x})}dx$ suddenly becomes much more manageable. Substituting our variables, we get:

By distributing the $\frac{1}{t}$ into the parenthesis, the expression cleans up beautifully:

Suddenly, the "monster" is gone. We are left with a rational function multiplied by an exponential. This is the moment where you must stop and look for a pattern.

The Hidden Derivative

This is the core of the problem. We are looking at an integral of the form $\int [\text{something}]e^{h(t)}dt$. Whenever you see this, you should immediately recall the product rule for differentiation:

In our case, the exponent is $h(t)=t+\frac{1}{t}$. Let us calculate its derivative: $h^{\prime }(t)=1-\frac{1}{t^2}$.

Now, look at our integrand again: $(t+2-\frac{1}{t}-\frac{1}{t^2})e^{(t+\frac{1}{t})}$. We need to find a function $f(t)$ such that its derivative plus $f(t)$ times $h^{\prime }(t)$ matches our polynomial.

This is where the JEE intuition kicks in. We need a function $f(t)$ that, when differentiated and combined with $h^{\prime }(t)$, yields our polynomial. Let us test the simplest candidate: $f(t)=t+1$.

The Verification

If $f(t)=t+1$, then $f^{\prime }(t)=1$. Let us plug this into our product rule template:

Expanding this, we get:

It is a perfect match! The polynomial in our integral is exactly the derivative of $(t+1)e^{(t+1/t)}$. The math has aligned perfectly.

The Final Victory

We are almost there. We have found that the integral is simply $(t+1)e^{(t+1/t)}+c$. Now, we must return to the world of $x$.

Substituting $t=e^x$ back into our result, we get:

Comparing this to the target form $g(x)e^{(e^x+e^{-x})}+c$, we can clearly see that $g(x)=e^x+1$. The final step is trivial but satisfying: finding $g(0)$.

And there you have it. What started as a terrifying integral was dismantled by a simple substitution and a keen eye for the product rule.

Remember, in JEE Advanced, the math is rarely trying to trick you with complexity; it is trying to reward you for your ability to see the underlying structure. Keep practicing this pattern recognition, and you will find that even the most intimidating integrals become nothing more than a puzzle waiting to be solved.