Evaluate the integral by making a substitution and then using a table of integrals.
-$\int \frac { e ^ { x } } { e ^ { 2 x } - 25 } d x$
A) $\frac { 1 } { 10 } \ln \left| \frac { 5 - \mathrm { e } ^ { 2 \mathrm { x } } } { \mathrm { e } ^ { 2 \mathrm { x } } + 5 } \right| + \mathrm { C }$
B) $\frac { 1 } { 10 } \ln \left| \frac { 5 - x } { x + 5 } \right| + C$
C) $\frac { 1 } { 10 } \ln \left| \frac { 5 - \mathrm { e } ^ { \mathrm { x } } } { \mathrm { e } ^ { \mathrm { x } } + 5 } \right| + \mathrm { C }$
D) $\frac { 1 } { 10 } \ln \left| \frac { \mathrm { e } ^ { \mathrm { x } } + 5 } { \mathrm { e } ^ { \mathrm { x } } - 5 } \right| + \mathrm { C }$

Question

Quizplus · Accepted Answer

Let $u=e^x$, then $du=e^x dx$. 
Substituting this into the integral: $$\int \frac{e^x}{e^{2x}-25} dx=\int\frac{1}{u^2-25}du$$ 
We recognize that this is a standard integral with the form: $$\int\frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C$$ 
Thus, we can substitute $x=u$ and $a=5$ to get:
$$\int \frac { e ^ { x } } { e ^ { 2 x } - 25 } d x=\frac{1}{10}\ln\left|\frac{u-5}{u+5}\right|+C=\frac{1}{10}\ln\left|\frac{e^x-5}{e^x+5}\right|+C$$ 
Therefore, the correct choice is C.

Evaluate the Integral by Making a Substitution and Then Using ∫exe2x−25dx\int \frac { e ^ { x } } { e ^ { 2 x } - 25 } d x∫e2x−25ex​dx

Evaluate the Integral by Making a Substitution and Then Using $\int \frac { e ^ { x } } { e ^ { 2 x } - 25 } d x$