Question about integral and natural log

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JamesGoh
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Homework Statement



Find the integral to the following expression

[itex]\int\frac{1}{x-1}dx[/itex]


Homework Equations



[itex]\int udv = uv - \int vdu[/itex]

[itex]\int\frac{1}{x}dx = ln(x)[/itex]

The Attempt at a Solution



Given the information above, would a correct answer to this problem be ln(x-1) ? If not, what am I doing wrong ?

thanks
 
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how non need to use int by parts a simple substitution will do, ln(x-1) is correct

if you don't show what you doing (your steps it would eb hard to tell what was going wrong as well ;)
 
lanedance said:
how non need to use int by parts a simple substitution will do, ln(x-1) is correct

if you don't show what you doing (your steps it would eb hard to tell what was going wrong as well ;)


oh i just realized if you use substitution of variable u=x-1, it makes it easy to show that integral of 1/(x-1) is ln(x-1) :)
 
Last edited:
JamesGoh said:

Homework Statement



Find the integral to the following expression

[itex]\int\frac{1}{x-1}dx[/itex]


Homework Equations



[itex]\int udv = uv - \int vdu[/itex]

[itex]\int\frac{1}{x}dx = ln(x)[/itex]

The Attempt at a Solution



Given the information above, would a correct answer to this problem be ln(x-1) ? If not, what am I doing wrong ?

thanks

You can attack the problem as follows

let F(x)=[itex]\int\frac{1}{x-1}dx[/itex]


From then on make substitution x=x+1

which yields

F(x+1)=[itex]\int\frac{1}{x}dx[/itex]

Find out this integral which is quite obvious ( do not forget to add a constant to it )

since what you found is F(x+1) now re-substitute x=x-1
 
lanedance said:
good guess, is that all you need?


yeah pretty much. i worked it out myself using the substituiton and it came out fine