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