Question
\( \int ^ { \infty } \frac { 1 } { x + 1 } \)
Ask by Daniel West. in Greenland
Feb 20,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The integral \(\int_{1}^{\infty} \frac{1}{x + 1} \, dx\) does not have a finite value; it diverges.
Solution
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
\(\int_{1}^{+\infty} \frac{1}{x+1} dx\)
- step1: Rewrite the improper integral:
\(\lim _{a\rightarrow +\infty}\left(\int_{1}^{a} \frac{1}{x+1} dx\right)\)
- step2: Evaluate the integral:
\(\lim _{a\rightarrow +\infty}\left(\ln{\left(\frac{\left|a+1\right|}{2}\right)}\right)\)
- step3: Rewrite the expression:
\(\ln{\left(\lim _{a\rightarrow +\infty}\left(\frac{\left|a+1\right|}{2}\right)\right)}\)
- step4: Calculate:
\(\ln{\left(+\infty\right)}\)
- step5: Calculate:
\(+\infty\)
- step6: Calculate:
\(\textrm{Diverges}\)
The integral
\[
\int_{1}^{\infty} \frac{1}{x + 1} \, dx
\]
is an improper integral. Upon evaluation, it diverges, meaning that it does not converge to a finite value.
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Simplify this solution Bonus Knowledge
The integral \( \int_{0}^{\infty} \frac{1}{x + 1} \, dx \) diverges. It approaches infinity as we take the upper limit to infinity. This can be intuitively understood by recognizing that \( \frac{1}{x + 1} \) behaves like \( \frac{1}{x} \) for large \( x \), which is known to diverge when integrated from 0 to infinity. If you're looking for a finite result, you might want to check limits or definite integrals instead. For example, evaluating \( \int_{0}^{a} \frac{1}{x + 1} \, dx \) leads to a logarithmic function, specifically \( \ln(a + 1) \), which gives a finite result as \( a \) approaches infinity.
