Question

determine whether the integral is convergent or divergent. if it is convergent, evaluate it. (if the quantity diverges, enter diverges.)
int_{1}^{infty}\frac{37}{ln(x)}dx

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please try again. first evaluate the integral from 1 to t. determine whether or not this is convergent as t approaches (infty). if it is, evaluate the limit of the integral.

Explanation:

Step1: Use substitution

Let $u = \ln(x)$, then $du=\frac{1}{x}dx$. When $x = 1$, $u=\ln(1) = 0$; when $x=t$, $u=\ln(t)$. The integral $\int_{1}^{t}\frac{37}{\ln(x)}\cdot\frac{1}{x}dx$ becomes $\int_{0}^{\ln(t)}37du$.

Step2: Evaluate the integral

$\int_{0}^{\ln(t)}37du=37u\big|_{0}^{\ln(t)}=37\ln(t)-37\times0 = 37\ln(t)$.

Step3: Find the limit as $t\to\infty$

We want to find $\lim_{t\to\infty}\int_{1}^{t}\frac{37}{x\ln(x)}dx=\lim_{t\to\infty}37\ln(t)$. As $t\to\infty$, $\ln(t)\to\infty$, so $37\ln(t)\to\infty$.

Answer:

DIVERGES