The value of $\smallint \frac{{{e^x}}}{{{e^{2x}} - 4}}dx$ will be ______, where C is an arbitrary constant.

Question

The value of $\smallint \frac{{{e^x}}}{{{e^{2x}} - 4}}dx$ will be ______, where C is an arbitrary constant.

TuteeHUB · Accepted Answer

$\frac{1}{2}\log \left| {\frac{{{e^{2x}} + 2}}{{{e^{2x}} - 2}}} \right| + C$

TuteeHUB · Answer

$\frac{1}{2}\log \left| {\frac{{{e^x} + 1}}{{{e^x} - 1}}} \right| + C$

TuteeHUB · Answer

$\frac{1}{3}\log \left| {\frac{{{2e^x} - 1}}{{{2e^x} + 1}}} \right| + C$

TuteeHUB · Answer

$\frac{1}{4}\log \left| {\frac{{{e^x} - 2}}{{{e^x} + 2}}} \right| + C$

1.	The value of \(\smallint \frac{{{e^x}}}{{{e^{2x}} - 4}}dx\) will be ______, where C is an arbitrary constant.
A.	\(\frac{1}{2}\log \left\| {\frac{{{e^x} + 1}}{{{e^x} - 1}}} \right\| + C\)
B.	\(\frac{1}{3}\log \left\| {\frac{{{2e^x} - 1}}{{{2e^x} + 1}}} \right\| + C\)
C.	\(\frac{1}{4}\log \left\| {\frac{{{e^x} - 2}}{{{e^x} + 2}}} \right\| + C\)
D.	\(\frac{1}{2}\log \left\| {\frac{{{e^{2x}} + 2}}{{{e^{2x}} - 2}}} \right\| + C\)
Answer» D. \(\frac{1}{2}\log \left\| {\frac{{{e^{2x}} + 2}}{{{e^{2x}} - 2}}} \right\| + C\)

The value of \(\smallint \frac{{{e^x}}}{{{e^{2x}} - 4}}dx\) will be ______, where C is an arbitrary constant.