Evaluate the improper integral or state that it is divergent.
-$$\int _ { 5 } ^ { \infty } \frac { d t } { t ^ { 2 } - 4 t }$$
A) $- \frac { 1 } { 4 } \ln 5$
B) $\frac { 1 } { 5 } \ln 4$
C) $\frac { 1 } { 4 } \ln 5$
D) $4 \ln 5$

Question

Quizplus · Accepted Answer

We can rewrite the integrand as $\frac{1}{t(t-4)}$. We notice that as $t$ approaches 4, the denominator approaches 0, so we have a vertical asymptote at $t=4$. Therefore, we should split the integral into two parts: $\int_5^4 \frac{dt}{t(t-4)}$ and $\int_4^\infty \frac{dt}{t(t-4)}$.

For the first integral, we can use partial fraction decomposition: $\frac{1}{t(t-4)} = \frac{1}{4t} - \frac{1}{4(t-4)}.$ Then, 
\begin{align*}
\int_5^4 \frac{dt}{t(t-4)} &= \left[\frac{1}{4}\ln|t| - \frac{1}{4}\ln|t-4|ight]_5^4 \
&= -\frac{1}{4}\ln 5.
\end{align*}

For the second integral, we can use partial fraction decomposition again: $\frac{1}{t(t-4)} = \frac{1}{4t-16} - \frac{1}{4t}$. Then, 
\begin{align*}
\int_4^\infty \frac{dt}{t(t-4)} &= \lim_{a 	o \infty} \left[\frac{1}{4}\ln|t-4| - \frac{1}{4}\ln|t|ight]_4^a \
&= \lim_{a 	o \infty} \left[\frac{1}{4}\ln\left|\frac{t-4}{t}ight|ight]_4^a \
&= \frac{1}{4}\ln\left|\frac{a-4}{a}ight|.
\end{align*}

Since this limit approaches $\frac{1}{4}\ln 1 = 0$ as $a$ goes to infinity, we can ignore this integral. Therefore, the entire integral evaluates to $-\frac{1}{4}\ln 5$.

Evaluate the Improper Integral or State That It Is Divergent ∫5∞dtt2−4t\int _ { 5 } ^ { \infty } \frac { d t } { t ^ { 2 } - 4 t }∫5∞​t2−4tdt​

Evaluate the Improper Integral or State That It Is Divergent $\int _ { 5 } ^ { \infty } \frac { d t } { t ^ { 2 } - 4 t }$