Animated Solution for Mathematics - Indefinite Integration: Let $I=\int \frac{e^x}{e^{4x}+e^{2x}+1}dx,J=\int \frac{e^{-x}}{e^{-4x}+e^{-2x}+1}dx$. Then, for an arbitrary constant $C$, the value of $J-I$ equals

Welcome, future engineers! Today, we are tackling a problem that often intimidates students at first glance. We are given two integrals:

At first, they look like distant cousins, but they are actually mirror images. The key to solving this is to realize that the negative exponents in $J$ are just an illusion.

Let us perform a little algebraic surgery on $J$. If we multiply both the numerator and the denominator by $e^{4x}$, we get:

Suddenly, the negative powers vanish, and we are left with:

Look at that! The denominator of $J$ is now identical to the denominator of $I$. This is the breakthrough we needed.

Now that the denominators are aligned, finding $J-I$ becomes a straightforward task. We can combine them into a single integral:

Notice the numerator: $e^{3x}-e^x$. We can factor out $e^x$ to get $e^x(e^{2x}-1)$.

This is a massive hint because the derivative of $e^x$ is $e^x$. This screams for a substitution. Let us set $t=e^x$, which implies $dt=e^{x}dx$. Our integral transforms into the algebraic form:

We have arrived at a form that is a favorite in JEE Advanced examinations. Whenever you see a quadratic expression in the numerator and a quartic expression in the denominator, the standard technique is to divide both the numerator and the denominator by $t^2$.

Let us do that:

Now, look at the denominator. We can rearrange it as $(t^2+\frac{1}{t^2})+1$. We know that $(t+\frac{1}{t}{)}^2=t^2+\frac{1}{t^2}+2$, which means $t^2+\frac{1}{t^2}=(t+\frac{1}{t}{)}^2-2$.

Substituting this back, our denominator becomes $(t+\frac{1}{t}{)}^2-2+1$, which simplifies to $(t+\frac{1}{t}{)}^2-1$.

We are almost there. Let us make our second substitution: $u=t+\frac{1}{t}$. Then, $du=(1-\frac{1}{t^2})dt$. This perfectly matches our numerator!

Our integral now becomes the simple, standard form:

Using the standard formula $\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+C$, with $a=1$, we get:

Finally, we back-substitute $u=t+\frac{1}{t}$ and then $t=e^x$. The expression becomes:

Multiplying the numerator and denominator inside the log by $t$, we get:

Substituting $t=e^x$, we arrive at our final answer:

This matches Option 3 perfectly. You have successfully navigated the complexity and emerged victorious!