lim In ix 1 lim 를 In (는) + tan Ih | In (x 1) + tan- x-, lim In () + tan-x %3D tan 3. Write out the first five terms of the

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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For this question. I'm confused why the answer is pi/4. I circled in green where I was trying to work it out at the end. I understand partial fraction decomp , just not understanding the end part how we got pi/4. I thought answer was infinity because infinity - a number is infinity.

∫₀∞ (1/(x + 1)√(x² + 1)) dx = lim (b → ∞) [ (1/2) ln|x + 1| - (1/2) ln(x² + 1) + (1/2) tan⁻¹x ]₀ᵇ 

= lim (b → ∞) [ (1/2) ln( (b + 1) / √(b² + 1) ) + (1/2) tan⁻¹b ] - [(1/2) ln(1/√1) + (1/2) tan⁻¹0] = (1/2) ln1 + (1/2) - (1/2) ln1 - (1/2) x 0 = π/4

3. Write out the first five terms of the sequence...
Transcribed Image Text:∫₀∞ (1/(x + 1)√(x² + 1)) dx = lim (b → ∞) [ (1/2) ln|x + 1| - (1/2) ln(x² + 1) + (1/2) tan⁻¹x ]₀ᵇ = lim (b → ∞) [ (1/2) ln( (b + 1) / √(b² + 1) ) + (1/2) tan⁻¹b ] - [(1/2) ln(1/√1) + (1/2) tan⁻¹0] = (1/2) ln1 + (1/2) - (1/2) ln1 - (1/2) x 0 = π/4 3. Write out the first five terms of the sequence...
**Transcription of Mathematical Notes for Educational Website**

---

Given problem:

\[ \int \frac{1}{x^2+1} \, dx \]

Substitution method:

1. **Substitution 1:**
   - Set \( u = x^2 + 1 \)
   - Then, \( du = 2x \, dx \)

2. **Substitution 2:**
   - Set \( u = x^2 \)
   - Then, \( du = 2x \, dx \)

---

**Integration Steps:**

\[ \int \frac{1}{2} \frac{1}{x^2+1} \, dx = \frac{1}{2} \left(\frac{-1}{x^2+1}\right) \left(-1\right) \]

\[ = \frac{-1}{2} \left( - \frac{1}{x^2+1} \right) \]

\[ = \frac{1}{2} \ln |x^2+1| + \text{const} \]

\[ = \frac{1}{2} \ln|x^2+1| + \tan^{-1}x \]

---

**Evaluating the Limits from \(-\infty\) to \(+\infty\):**

---

(Calculation is within a green boxed region)

1. Limit as \( x \to +\infty \):

\[ \int_{0}^{+\infty} \frac{1}{x^2+1} \, dx = \frac{1}{2} \ln | +\infty + 1|  \]

\[ = \frac{1}{2} \ln \infty = \text{undefined} \]

2. Limit as \( x \to 0 \):

\[ \int_{0}^{0} \frac{1}{x^2+1} \, dx = 0 \]

3. Limit as \( x \to -\infty \):

\[ \int_{0}^{-\infty} \frac{1}{x^2+1} \, dx = \frac{1}{2} \ln | -\infty + 1 | \]

\[ = \frac{1}{2} \ln \infty = \text{undefined} \]

Computation yields results which suggest careful handling of limits,
Transcribed Image Text:**Transcription of Mathematical Notes for Educational Website** --- Given problem: \[ \int \frac{1}{x^2+1} \, dx \] Substitution method: 1. **Substitution 1:** - Set \( u = x^2 + 1 \) - Then, \( du = 2x \, dx \) 2. **Substitution 2:** - Set \( u = x^2 \) - Then, \( du = 2x \, dx \) --- **Integration Steps:** \[ \int \frac{1}{2} \frac{1}{x^2+1} \, dx = \frac{1}{2} \left(\frac{-1}{x^2+1}\right) \left(-1\right) \] \[ = \frac{-1}{2} \left( - \frac{1}{x^2+1} \right) \] \[ = \frac{1}{2} \ln |x^2+1| + \text{const} \] \[ = \frac{1}{2} \ln|x^2+1| + \tan^{-1}x \] --- **Evaluating the Limits from \(-\infty\) to \(+\infty\):** --- (Calculation is within a green boxed region) 1. Limit as \( x \to +\infty \): \[ \int_{0}^{+\infty} \frac{1}{x^2+1} \, dx = \frac{1}{2} \ln | +\infty + 1| \] \[ = \frac{1}{2} \ln \infty = \text{undefined} \] 2. Limit as \( x \to 0 \): \[ \int_{0}^{0} \frac{1}{x^2+1} \, dx = 0 \] 3. Limit as \( x \to -\infty \): \[ \int_{0}^{-\infty} \frac{1}{x^2+1} \, dx = \frac{1}{2} \ln | -\infty + 1 | \] \[ = \frac{1}{2} \ln \infty = \text{undefined} \] Computation yields results which suggest careful handling of limits,
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