Evaluate the integral.
-$$\int \frac { e ^ { t } d t } { e ^ { 2 t } - 6 e ^ { t } + 5 }$$
A) $\frac { 1 } { 4 } \ln \left| \mathrm { e } ^ { \mathrm { t } } - 5 \right| - \frac { 1 } { 4 } \ln \left| \mathrm { e } ^ { \mathrm { t } } - 1 \right| + \mathrm { C }$
B) $\frac { 1 } { 4 } \ln \left| \mathrm { e } ^ { \mathrm { t } } - 5 \right| + \frac { 1 } { 4 } \ln \left| \mathrm { e } ^ { \mathrm { t } } - 1 \right| + \mathrm { C }$
C) $\frac { 1 } { 4 } \ln | t - 5 | - \frac { 1 } { 4 } \ln | t - 1 | + C$
D) $\frac { 1 } { 4 } \mathrm { e } ^ { \mathrm { t } } \ln \left| \mathrm { e } ^ { \mathrm { t } } - 5 \right| - \frac { 1 } { 4 } \mathrm { e } ^ { \mathrm { t } } \ln \left| \mathrm { e } ^ { \mathrm { t } } - 1 \right| + \mathrm { C }$

Question

Quizplus · Accepted Answer

The integral can be solved by partial fraction decomposition. The denominator $e^{2t} - 6e^t + 5$ can be factored as $(e^t - 5)(e^t - 1)$. Using partial fractions, we can write the integrand as a sum of fractions whose denominators are these factors. The solution involves finding constants A and B such that $\frac{e^t}{(e^t - 5)(e^t - 1)} = \frac{A}{e^t - 5} + \frac{B}{e^t - 1}$. Solving for A and B, integrating each term separately, and simplifying leads to the answer in option A.

Evaluate the Integral ∫etdte2t−6et+5\int \frac { e ^ { t } d t } { e ^ { 2 t } - 6 e ^ { t } + 5 }∫e2t−6et+5etdt​

Evaluate the Integral $\int \frac { e ^ { t } d t } { e ^ { 2 t } - 6 e ^ { t } + 5 }$