Question

$\int \frac{e^{3x}+e^x}{e^{4x}-e^{2x}+1}dx=$

• A. $\frac{1}{4}\log (e^{4x}-e^{2x}+1)+c$
• B. ${\tan }^{-1}(e^x-e^{-x})+c$
• C. ${\tan }^{-1}(e^x+e^{-x})+c$
• D. ${\tan }^{-1}(e^{-x}-e^x)+c$

Answer 1

The correct option is B ${\tan }^{-1}(e^x-e^{-x})+c$

$\int \left[\frac{e^{3x}+e^x}{e^{4x}-e^{2x}+1}\right]dx$

$\int \frac{e^{2x}(e+\frac{1}{e^x})}{(e^{4x}-e^{2x}+1)}dx=\int \left[\frac{e^x+\frac{1}{e^x}}{e^{2x}+\frac{1}{e^{2x}}+1}\right]dx$

$e^x-\frac{1}{e^x}=t(e^x+\frac{1}{e^x})dx=dt$

$\int \frac{dt}{t^2+1}=\int \frac{dt}{t^2+1}$

$\frac{1}{1}{\tan }^{-1}\left(\frac{t}{1}\right)+c$

${\tan }^{-1}(e^x-e^{-x})+c$