Substitution Method

  • #1
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:
 

Answers and Replies

  • #2
2,985
15
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:
 
  • #3
23
0
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:
  • #4
2,985
15
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.
 
  • #5
970
3

[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]

 
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  • #6
23
0
how do u decide that u is the positive or negative?

[tex] u=? +\sqrt{x+5} \ or \ u=?-\sqrt{x+5}[/tex]
 
  • #7
970
3

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]
 
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