Solve S (x+5)½/x-4 dx with Substitution Method

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Discussion Overview

The discussion revolves around solving the integral S (x+5)½/x-4 dx using the substitution method. Participants explore various substitution techniques and express differing opinions on the best approach to take, including the use of trigonometric substitution.

Discussion Character

  • Homework-related, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant expresses difficulty with the substitution method, suggesting it complicates the problem rather than simplifying it.
  • Another participant suggests using trigonometric substitution, questioning whether the original poster has attempted the problem before seeking help.
  • A participant proposes setting u² = x + 5, leading to an integral that involves partial fractions and long division, ultimately arriving at a similar answer to the original poster.
  • There is a discussion about the choice of the positive or negative square root when substituting u = ±√(x + 5), with participants questioning how to determine which value to use.
  • One participant provides a detailed solution and expresses confidence in their approach, while also referencing a mathematical identity related to hyperbolic functions.
  • Another participant reiterates the question of how to decide between the positive and negative roots in the context of the problem.
  • A later reply notes that there are two possible solutions in different quadrants, but concludes that the specific solution being discussed has a unique x-intercept, suggesting a preference for the positive root.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best substitution method to use, and there is ongoing debate about the choice of positive or negative roots in the substitution process.

Contextual Notes

Participants express uncertainty regarding the implications of choosing the positive or negative root, and the discussion includes various mathematical steps that remain unresolved.

Cradle_of_Knowledge
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Hi,

I have another problem about substitution Method. I think this method is used to make the problem to solve in easy way but it is making my procedure too long for this problem. Can you solve it by substitution method.

S (x+5)½/x-4 dx

where S is the sign of integral. The answer of this problem is

2(x+5)½+3ln[(x+5)½-3/(x+5)½+3]+c
:confused:
 
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You will have to do a trig substitution using either sin or cos. I hope your at least attemping these homework problems before you ask for the anwser, because they seem like homework. :rolleyes:
 
i set [tex]u^2= x+5[/tex]

i get integral:
[tex]2\int \frac {\2{u^2}}{(u+3)(u-3)} du[/tex]

then do the long division, do partial fractions then i get the same answer as u posted.

answer is : [tex]2u -3\ln(u+3) + 3\ln(u-3) + c[/tex]

i was wondering how do u know if
[tex]u=? +\sqrt{x+5} \ or \ u=?-\sqrt{x+5}[/tex]
?
 
Last edited:
nice choice of subsitution unggio, i take back having to use trig now that I see what you wrote.

You can learn to use the [tex]LaTeX[/tex] in the physics forum.
 

[tex]u^2 = x + 5[/tex]

[tex]2 \int \frac{u^2}{(u + 3)(u - 3)} du = 2 \left( u + \frac{3}{2} \ln [u - 3] - \frac{3}{2} \ln [u + 3] \right) + C[/tex]

[tex]2 \left( u + \frac{3}{2} \ln [u - 3] - \frac{3}{2} \ln [u + 3] \right) + C= 2u + 3 \ln (u - 3) - 3 \ln (u + 3) + C[/tex]

[tex]2u + 3 \ln (u - 3) - 3 \ln (u + 3) = 2 \sqrt{x + 5} + 3 \ln (\sqrt{x+5} - 3) - 3 \ln (\sqrt{x + 5} + 3) + C[/tex]

[tex]\boxed{ \int \frac{\sqrt{x + 5}}{x - 4} dx = 2 \sqrt{x + 5} + 3 \ln (\sqrt{x+5} - 3) - 3 \ln (\sqrt{x + 5} + 3) + C}[/tex]

My research also located an Identity:
[tex]\tanh^{-1} \left( \frac{u}{3} \right) = \frac{1}{2} ( \ln [u + 3] - \ln [u - 3] )[/tex]

 
Last edited:
how do u decide that u is the positive or negative?

[tex]u=? +\sqrt{x+5} \ or \ u=?-\sqrt{x+5}[/tex]
 

The functional identity:
[tex]\tanh^{-1} \left( \frac{u}{3} \right) = \frac{1}{2} ( \ln [u + 3] - \ln [u - 3] )[/tex]

There are two possible solutions in Quadrants I and III, therefore [tex]u = \pm \sqrt{x + 5}[/tex]

However, the solution:
[tex]2 \sqrt{x + 5} + 3 \ln (\sqrt{x+5} - 3) - 3 \ln (\sqrt{x + 5} + 3) + C = 0[/tex]

Has x-intercept only at [tex]x = 7.953[/tex], therefore [tex]u = + \sqrt{x + 5}[/tex]
 
Last edited:

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