91Ó°ÊÓ

We are given two integrals,

${∫}_1^{∞}\frac{1}{x+1}dx$And

${∫}_1^{∞}\frac{1}{x}dx$

Observe that $x+1>x$for all $xâ‰\yen 1$.

Then, $\frac{1}{x+1}<\frac{1}{x}$for all $xâ‰\yen 1$.

The comparison test for improper integrals states is given by,

Suppose f and g are two continuous functions on an interval I.

The improper integral of f on I also diverges, if the improper integral of g on I diverges and $0≤g\left(x\right)≤f\left(x\right)$for all $x∈I$.

The comparison test fails to tell about the convergence or divergence of ${∫}_1^{∞}\frac{1}{x+1}dx$based on the divergence of ${∫}_1^{∞}\frac{1}{x}dx$.

Observe that the graph of $y=\frac{1}{x+1}$is just a horizontal translation of the graph of $y=\frac{1}{x}$.

The area under the graph of $\frac{1}{x}$seems to grow without bound as $x→∞$.

Similarly, the area under the graph of $\frac{1}{x+1}$seems to grow without bound as $x→∞$but in a slower rate as compared to that of $\frac{1}{x}$.

Hence, ${∫}_1^{∞}\frac{1}{x+1}dx$diverges since the graph of ${∫}_1^{∞}\frac{1}{x}dx$diverges.