Question

If $I_1 = \int \frac{e^x}{e^{4x}+e^{2x}+1}dx$, $I_2 = \int \frac{e^{-x}}{e^{-4x}+e^{-2x}+1}dx$, then $I_2-I_1=$

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

Hint:

Integrals of the form $\int \frac{x^2\pm 1}{x^4+kx^2+1}dx$ can often be solved by dividing the numerator and denominator by $x^2$ and then substituting $u=x\mp 1/x$. This technique creates a simpler integral in terms of $u$.

Answer 1

The Correct Option is D

Step 1: Simplify the integrand of $I_2$.
The integrand of $I_2$ is $\frac{e^{-x}}{e^{-4x}+e^{-2x}+1}$. Multiply the numerator and denominator by $e^{4x}$: \[ \frac{e^{-x} \cdot e^{4x}}{(e^{-4x}+e^{-2x}+1) \cdot e^{4x}} = \frac{e^{3x}}{1+e^{2x}+e^{4x}}. \] So, $I_2 = \int \frac{e^{3x}}{e^{4x}+e^{2x}+1}dx$.

Step 2: Set up the integral for $I_2 - I_1$.
\[ I_2 - I_1 = \int \frac{e^{3x}}{e^{4x}+e^{2x}+1}dx - \int \frac{e^x}{e^{4x}+e^{2x}+1}dx = \int \frac{e^{3x}-e^x}{e^{4x}+e^{2x}+1}dx. \] Factor out $e^x$ from the numerator: \[ I_2 - I_1 = \int \frac{e^x(e^{2x}-1)}{e^{4x}+e^{2x}+1}dx. \]

Step 3: Perform a substitution.
Let $u = e^x$, so $du = e^x dx$. The integral becomes: \[ I_2 - I_1 = \int \frac{u^2-1}{u^4+u^2+1}du. \] Divide the numerator and denominator by $u^2$: \[ \int \frac{1-1/u^2}{u^2+1+1/u^2}du = \int \frac{1-1/u^2}{(u+1/u)^2-1}du. \]

Step 4: Perform a second substitution.
Let $v = u+\frac{1}{u}$. Then $dv = (1-\frac{1}{u^2})du$. The integral becomes: \[ \int \frac{1}{v^2-1}dv. \] This is a standard integral: $\frac{1}{2}\log\left|\frac{v-1}{v+1}\right|+c$.

Step 5: Substitute back to the original variable.
First, substitute back for $v$: \[ \frac{1}{2}\log\left|\frac{(u+1/u)-1}{(u+1/u)+1}\right|+c = \frac{1}{2}\log\left|\frac{u^2-u+1}{u^2+u+1}\right|+c. \] Next, substitute back $u=e^x$: \[ \frac{1}{2}\log\left(\frac{e^{2x}-e^x+1}{e^{2x}+e^x+1}\right)+c. \] To match the form in option (D), divide the numerator and denominator inside the log by $e^x$: \[ \frac{1}{2}\log\left(\frac{e^x-1+e^{-x}}{e^x+1+e^{-x}}\right)+c = \frac{1}{2}\log\left(\frac{e^x+e^{-x}-1}{e^x+e^{-x}+1}\right)+c. \]