Answered: 3x2 + 5 dx. x2 – 3x +2 Evaluate

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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Step 1

$\begin{array}{l} Integrate \\ \int \frac{3 x^{2} + 5}{x^{2} - 3 x + 2} d x \end{array}$

Step 2

$\begin{array}{l} Using long division \\ \int (\frac{9 x - 1}{x^{2} - 3 x + 2} + 3) d x \\ \int \frac{9 x - 1}{x^{2} - 3 x + 2} d x + \int 3 d x ....... (1) \\ Write \\ 9 x - 1 = \frac{9}{2} (2 x - 3) + \frac{25}{2} \\ = \int \frac{9 x - 1}{x^{2} - 3 x + 2} d x + 3 x \\ = \frac{9}{2} \int \frac{2 x - 3}{x^{2} - 3 x + 2} d x + \frac{25}{2} \int \frac{1}{x^{2} - 3 x + 2} d x ......... (2) \\ Now \\ Let \\ I_{1} = \int \frac{2 x - 3}{x^{2} - 3 x + 2} d x {and I}_{2} = \int \frac{1}{x^{2} - 3 x + 2} d x \\ {Now solve I}_{1} {and I}_{2} \\ I_{1} = \int \frac{2 x - 3}{x^{2} - 3 x + 2} d x \\ Substitute \\ w = x^{2} - 3 x + 2 \\ \frac{d w}{d x} = 2 x - 3 \\ d x = \frac{1}{2 x - 3} d w \\ {Put in I}_{1} \\ I_{1} = \int \frac{1}{w} d w \\ I_{1} = \ln (w) \\ Substitute w= x^{2} - 3 x + 2 \\ I_{1} = \ln (x^{2} - 3 x + 2) \\ {Now solve I}_{2} \\ I_{2} = \int \frac{1}{x^{2} - 3 x + 2} d x \\ Factorization \\ = \int \frac{1}{(x - 2) (x - 1)} d x \\ Partial fraction decomposition \\ = \int \frac{1}{x - 2} d x - \int \frac{1}{x - 1} d x \\ I_{2} = \ln (x - 2) + \ln (x - 1) \end{array}$