Question

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 (A) $\frac{1}{2}\log \left(\frac{e^{4x}-e^{2x}+1}{e^{4x}+e^{2x}+1}\right)+C$ (B) $\frac{1}{2}\log \left(\frac{e^{2x}+e^x+1}{e^{2x}-e^x+1}\right)+C$ (C) $\frac{1}{2}\log \left(\frac{e^{2x}-e^x+1}{e^{2x}+e^x+1}\right)+C$ (D) $\frac{1}{2}\log \left(\frac{e^{4x}+e^{2x}+1}{e^{4x}-e^{2x}+1}\right)+C$

Short answer

None of the given options correctly represent the value of $J-I$.

Step-by-step solution

- Rewrite the integrand of J

To make the integration of J easier, perform a substitution by letting u = -x. This changes the differential from dx to du and simplifies the integrand: $J=\int \frac{e^u}{e^{4u}+e^{2u}+1}du$.

- Recognize the similarity

Observe that the integrand of J is now identical to the integrand of I, but with u replacing x: $I=\int \frac{e^x}{e^{4x}+e^{2x}+1}dx$ and $J=\int \frac{e^u}{e^{4u}+e^{2u}+1}du$.

- Calculate I - J

Since I is a function in terms of x and J is the same function in terms of -x, we find the difference as $J-I=\int \frac{e^u}{e^{4u}+e^{2u}+1}du-\int \frac{e^x}{e^{4x}+e^{2x}+1}dx$. We can now reverse the substitution for J, replacing u with -x: $J-I=\int \frac{e^{-x}}{e^{-4x}+e^{-2x}+1}(-dx)-\int \frac{e^x}{e^{4x}+e^{2x}+1}dx$, which simplifies to $J-I=-\int \frac{e^{-x}}{e^{-4x}+e^{-2x}+1}dx-I$.

- Integrate and combine terms

Since J is the negative of I when substituting u back to x, we combine the terms: $J-I=-I-I=-2I$. To evaluate I, it requires noticing a pattern or performing a clever substitution that you can relate back to your original integrals; however, those exact methods are beyond the scope of this problem since we are to find J - I, not I itself.

- Conclude based on options given

By elimination, option (B) can only be correct if J = -I; however, since we have established that J - I = -2I, and none of the answer choices are equivalent to -2I, none of them are correct. Thus, if the options do not include a non-listed one, all the presented options are incorrect for the solution of J - I.

Key concepts

Exponential Functions

Exponential functions are a foundational concept in math that students encounter in various disciplines, including calculus. An exponential function is defined by the equation $f(x)=a^x$, where $a$ is a constant base and $x$ is the exponent. These functions are distinctive for their rapid growth or decay properties.

When studying exponential functions within the realm of integral calculus, it's crucial to recognize how the base of the exponential, typically $e$, the mathematical constant approximately equal to 2.71828, affects integration. The unique property of $e^x$ is that it remains unchanged after differentiation or integration, which is a pivotal aspect of solving integrals that include exponential terms.

The problem presented features integrals of fractions with exponential functions in the denominator. These are not plain exponential functions, rather they are rationals, reflecting a more complex application of exponential behaviors in calculus.

Integration Techniques

In integral calculus, selecting the right integration techniques can transform a seemingly complicated problem into a manageable one. Common integration methods include substitution, integration by parts, partial fractions, and trigonometric substitution. Each technique has its specific use case depending on the form of the function you're integrating.

In the exercise provided, the substitution technique was used to tackle the integrals. By letting $u=-x$, the second integral $J$ was transformed to have the same integrand as $I$, but in terms of $u$ instead of $x$. Substitution is especially powerful when you recognize a part of the integrand that can simplify the integral if it were expressed in different terms. In the case of exponential functions, it's often helpful to perform substitutions to reduce the integrand to a familiar form, making it easier to integrate.

Indefinite Integrals

Indefinite integrals, also known as antiderivatives, are integrals that do not specify limits of integration. In essence, finding an indefinite integral means finding a function whose derivative matches the given function. The distinction between indefinite and definite integrals lies in their application; indefinite integrals are about finding a general form, while definite integrals calculate the area under the curve within specific bounds.

In the context of the exercise, both $I$ and $J$ are indefinite integrals as no limits of integration are provided. When computing $J-I$, we focus on the algebraic manipulation of the indefinite integrals and not on calculating a numerical value. It's essential to remember that an indefinite integral will include an arbitrary constant $C$, because differentiation of a constant yields zero and cannot be determined from the integral alone.