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Evaluate the integral √ (k+ 2) ( k + 3) dx

Evaluate the integral √ (k+ 2) ( k + 3) dx

Calculus: Early Transcendentals
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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**Evaluate the Integral** \[ \int \frac{12}{(k+2)(k+3)} \, dk \]
Transcribed Image Text:**Evaluate the Integral** \[ \int \frac{12}{(k+2)(k+3)} \, dk \]
--- ### Integral Evaluation Using Trigonometric Substitution #### Problem Statement: **Evaluate the following integrals using a trigonometric substitution. Hint: complete the square for part b.** ##### a) Integral of (x^2 / (1 + x^2)) dx ##### b) Integral of sqrt(x^2 + 6x) dx (Note: A clever way to evaluate the first integral is to add and subtract one to the numerator or use long division. No points will be given for these approaches.) #### Solution for Part a: \[ \int \frac{x^2}{1 + x^2} \, dx \] The given integral can be simplified as: \[ \int \frac{x^2 + 1 - 1}{1 + x^2} \, dx = \int \left( \frac{1 + x^2}{1 + x^2} - \frac{1}{1 + x^2} \right) \, dx \] \[ = \int \left( 1 - \frac{1}{1 + x^2} \right) \, dx \] \[ = \int 1 \, dx - \int \frac{1}{1 + x^2} \, dx \] The integral of 1 with respect to x is x, and the integral of \( \frac{1}{1 + x^2} \) is \( \tan^{-1} x\): \[ \int 1 \, dx = x \] \[ \int \frac{1}{1 + x^2} \, dx = \tan^{-1} x \] Thus, the solution is: \[ x - \tan^{-1} x + C \] Where C is the constant of integration. ---
Transcribed Image Text:--- ### Integral Evaluation Using Trigonometric Substitution #### Problem Statement: **Evaluate the following integrals using a trigonometric substitution. Hint: complete the square for part b.** ##### a) Integral of (x^2 / (1 + x^2)) dx ##### b) Integral of sqrt(x^2 + 6x) dx (Note: A clever way to evaluate the first integral is to add and subtract one to the numerator or use long division. No points will be given for these approaches.) #### Solution for Part a: \[ \int \frac{x^2}{1 + x^2} \, dx \] The given integral can be simplified as: \[ \int \frac{x^2 + 1 - 1}{1 + x^2} \, dx = \int \left( \frac{1 + x^2}{1 + x^2} - \frac{1}{1 + x^2} \right) \, dx \] \[ = \int \left( 1 - \frac{1}{1 + x^2} \right) \, dx \] \[ = \int 1 \, dx - \int \frac{1}{1 + x^2} \, dx \] The integral of 1 with respect to x is x, and the integral of \( \frac{1}{1 + x^2} \) is \( \tan^{-1} x\): \[ \int 1 \, dx = x \] \[ \int \frac{1}{1 + x^2} \, dx = \tan^{-1} x \] Thus, the solution is: \[ x - \tan^{-1} x + C \] Where C is the constant of integration. ---
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