Evaluate $\int \frac{6 e^{-2 w}}{1-5 e^{-2 w}} d w$ .
A) $\frac{12 e^{-2 w}}{1 t-10 e^{-2 w}}+C$
B) $\frac{6 e^{-2 w}}{2 t-5 e^{-2 w}}+C$
C) $\frac{3}{5} \ln \left|1-5 e^{-2 w}\right|+C$
D) $\frac{1}{10} \ln \left|1-5 e^{-2 w}\right|+C$

Question

Quizplus · Accepted Answer

The integral can be solved by recognizing it as a standard form of an integral involving a natural logarithm. Let $u = 1 - 5e^{-2w}$, then $du = 10e^{-2w} dw$. The integral becomes $\int \frac{6}{10} \frac{1}{u} du$, which simplifies to $\frac{3}{5} \ln |u| + C$, or $\frac{3}{5} \ln |1-5e^{-2w}| + C$.

Evaluate ∫6e−2w1−5e−2wdw\int \frac{6 e^{-2 w}}{1-5 e^{-2 w}} d w∫1−5e−2w6e−2w​dw A) 12e−2w1t−10e−2w+C\frac{12 e^{-2 w}}{1 t-10 e^{-2 w}}+C1t−10e−2w12e−2w​+C

Evaluate $\int \frac{6 e^{-2 w}}{1-5 e^{-2 w}} d w$
A) $\frac{12 e^{-2 w}}{1 t-10 e^{-2 w}}+C$