Use a table of integrals with forms involving the trigonometric functions to find the integral. $\int \frac { 1 } { 1 + e ^ { 5 x } } d x$
A) $\int \frac { 1 } { 1 + e ^ { 5 x } } d x = x ^ { 2 } - \frac { 1 } { 5 } \log \left( e ^ { 5 x } + 1 \right) + C$
B) $\int \frac { 1 } { 1 + e ^ { 5 x } } d x = x - \log \left( e ^ { 5 x } + 1 \right) + C$
C) $\int \frac { 1 } { 1 + e ^ { 5 x } } d x = \log \left( e ^ { 5 x } + 1 \right) + C$
D) $\int \frac { 1 } { 1 + e ^ { 5 x } } d x = x - \frac { 1 } { 5 } \log \left( e ^ { 5 x } + 1 \right) + C$
E) $\int \frac { 1 } { 1 + e ^ { 5 x } } d x = \frac { 1 } { 5 } \log \left( e ^ { 5 x } + 1 \right) + C$

Question

Quizplus · Accepted Answer

The integral can be solved using substitution. Let $ u = e^{5x} $, then $ du = 5e^{5x} dx $ or $ dx = \frac{du}{5u} $. The integral becomes $\int \frac{1}{1+u} \cdot \frac{1}{5u} du$, which simplifies to $\frac{1}{5} \int \frac{1}{u(1+u)} du$. This can be solved using partial fraction decomposition, leading to the result $ x - \frac{1}{5} \log(e^{5x} + 1) + C $.

Use a Table of Integrals with Forms Involving the Trigonometric ∫11+e5xdx\int \frac { 1 } { 1 + e ^ { 5 x } } d x∫1+e5x1​dx

Use a Table of Integrals with Forms Involving the Trigonometric $\int \frac { 1 } { 1 + e ^ { 5 x } } d x$