number 7 For problems 7 & 8, give a power series representation for the integral of the following function.7. $ h(x) = \frac{7x}{3 - 6x} $

Name: number 7 For problems 7 & 8, give a power series representation for th
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Description: number 7 For problems 7 & 8, give a power series representation for the integral of the following function.7. $ h(x) = \frac{7x}{3 - 6x} $

Given, hx=7x3-6x ∫hx=7x∫13-6x dx-∫d7xdx∫3-6x-1 dx=7x∫131-2xdx-∫d7xdx∫…

Answered: 7x 7. h (ӕ) 3 — 6x

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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number 7

For problems 7 & 8, give a power series representation for the integral of the following function.

7. $ h(x) = \frac{7x}{3 - 6x} $

Expert Solution

Step 1

Given,

$h (x) = \frac{7 x}{3 - 6 x}$

$\begin{array}{rcl} \int h (x) & = & 7 x \int \frac{1}{3 - 6 x} d x - \int \frac{d (7 x)}{d x} \int {(3 - 6 x)}^{- 1} d x \\ = & 7 x \int \frac{1}{3 (1 - 2 x)} d x - \int \frac{d (7 x)}{d x} \int \frac{1}{3 (1 - 2 x)} d x \\ = & \frac{7 x}{3} \frac{\ln |1 - 2 x|}{- 2} - \frac{7}{3} \int \frac{\ln |1 - 2 x|}{- 2} \\ = & \frac{- 7 x}{6} (\ln |1 - 2 x|) + \frac{7}{6} \int \ln |1 - 2 x| d x \end{array}$

$\begin{array}{rcl} = \end{array} \begin{array}{rcl} \frac{- 7 x}{6} \end{array} \begin{array}{rcl} (\ln |1 - 2 x|) \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} \frac{7}{6 (- 2)} \end{array} \begin{array}{rcl} (1 - 2 x) \end{array} \begin{array}{rcl} (\ln |1 - 2 x|) \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} \frac{7}{6 (- 2)} \end{array} \begin{array}{rcl} |1 - 2 x| \end{array}$

$\begin{array}{rcl} \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} \frac{7}{12} \end{array} \begin{array}{rcl} (\ln |1 - 2 x|) \end{array} \begin{array}{rcl} (2 - 1 + 2 x) \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} \frac{7}{12} \end{array} \begin{array}{rcl} |1 - 2 x| \end{array}$

$\begin{array}{rcl} \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} \frac{7}{12} \end{array} \begin{array}{rcl} (\ln |1 - 2 x|) \end{array} \begin{array}{rcl} (1 + 2 x) \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} \frac{7}{12} \end{array} \begin{array}{rcl} |1 - 2 x| \end{array}$

$= \frac{7}{12} (\ln |1 - 2 x| (1 + 2 x) + |1 - 2 x|)$

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