**Partial Fraction Expansion and Integration**The partial fraction expansion of \[\frac{5x + 18x^2 + 13x - 20}{(x+3)(x^2 + 2x + 5)}\] is:\[\frac{2x - 5}{x^2 + 2x + 5} + \frac{3}{x+3} - \frac{4}{(x+3)^2}.\]Use this information to evaluate:\[\int \frac{5x + 18x^2 + 13x - 20}{(x+3)(x^2 + 2x + 5)} \, dx.\]

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Description: **Partial Fraction Expansion and Integration**The partial fraction expansion of \[\frac{5x + 18x^2 + 13x - 20}{(x+3)(x^2 + 2x + 5)}\] is:\[\frac{2x - 5}{x^2 + 2x + 5} + \frac{3}{x+3} - \frac{4}{(x+3)^2}.\]Use this information to evaluate:\[\int \fra

Given Data: The partial fraction expansion is: 5x+18x2+13x-20x+32x2+x+5=2x-5x2+2x+5+3x+3-4x+32…

Answered: 5x+ 18x2+13r -20 is: (x+3) (x²+2x+5)…

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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**Partial Fraction Expansion and Integration**

The partial fraction expansion of

\[
\frac{5x + 18x^2 + 13x - 20}{(x+3)(x^2 + 2x + 5)}
\]

is:

\[
\frac{2x - 5}{x^2 + 2x + 5} + \frac{3}{x+3} - \frac{4}{(x+3)^2}.
\]

Use this information to evaluate:

\[
\int \frac{5x + 18x^2 + 13x - 20}{(x+3)(x^2 + 2x + 5)} \, dx.
\]

Expert Solution

Step 1

Given Data:

The partial fraction expansion is: $\frac{5 x + 18 x^{2} + 13 x - 20}{{(x + 3)}^{2} (x^{2} + x + 5)} = \frac{2 x - 5}{x^{2} + 2 x + 5} + \frac{3}{x + 3} - \frac{4}{{(x + 3)}^{2}}$

Integral of the first term is,

$\begin{array}{rcl} \int \frac{2 x - 5}{x^{2} + 2 x + 5} d x & = & \int \frac{(2 x + 2) - 7}{x^{2} + 2 x + 5} d x \\ = & \int \frac{2 x + 2}{x^{2} + 2 x + 5} d x - 7 \int \frac{1}{x^{2} + 2 x + 5} d x \\ = & \int \frac{2 x + 2}{x^{2} + 2 x + 5} d x - 7 \int \frac{1}{4 + {(x + 1)}^{2}} d x \\ = & \int \frac{2 x + 2}{x^{2} + 2 x + 5} d x - 7 \int \frac{1}{{(2)}^{2} + {(x + 1)}^{2}} d x \\ = & \ln |x^{2} + 2 x + 5| - \frac{7}{2} (\tan^{- 1} (\frac{x + 1}{2})) + c \int \frac{1}{{(a)}^{2} + u^{2}} d u = \frac{1}{a} \tan^{- 1} \frac{u}{a} + c \end{array}$

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