Question about integral and natural log

[JamesGoh]

Homework Statement

Find the integral to the following expression

\int\frac{1}{x-1}dx

Homework Equations

\int udv = uv - \int vdu

\int\frac{1}{x}dx = ln(x)

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

 

[lanedance]

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 woudl eb hard to tell what was going wrong as well ;)

 

[JamesGoh]

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 woudl 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) :)

 

[lanedance]

good guess, is that all you need?

 

[stallionx]

Homework Statement

Find the integral to the following expression

\int\frac{1}{x-1}dx

Homework Equations

\int udv = uv - \int vdu

\int\frac{1}{x}dx = ln(x)

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)=\int\frac{1}{x-1}dx


From then on make substitution x=x+1

which yields

F(x+1)=\int\frac{1}{x}dx

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

 

[JamesGoh]

good guess, is that all you need?

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

 

[Hootenanny]

From then on make substitution x=x+1

 

[lanedance]

agree, would always use a different variable to represent the substitution eg. u = x-1, then du = dx