Question #913e7

2 Answers
Rhys
Jan 8, 2018

# x + ln|x-1| +c #

Explanation:

We can use a simple #u# - substitution...

Let # u = x-1 #

#=> u +1 = x #

#=> du = dx #

#=> int x/(x-1) dx = int (u+1)/u du #

#=> -= int u/u + 1/u du -= int 1 +1/u du #

#=> u + ln|u| + c #

Substituting back:

#=> x-1 + ln|x-1| +c_0# as #color(blue)(c_0 -1 # is just another constant...

# => color(red)( x + ln|x-1| + c ) #

#c in RR # - Just a constant

Alvin L.
Jan 8, 2018

#int\ x/(x-1)\ dx=x+ln|x-1|+C#

Explanation:

Alternatively, we can do a bit of a trick. If we use the fact that #x-1+1# still equals #x#, we can rewrite the integral:
#int\ x/(x-1)\ dx=int\ (x-1+1)/(x-1)\ dx=int\ (x-1)/(x-1)+1/(x-1)\ dx#

#=int\ 1+1/(x-1)\ dx=x+ln|x-1|+C#

As long as you're comfortable with doing the u-substitution of #1/(x-1)# in your head, this method will probably get you the answer quite a bit quicker.

Related questions
Impact of this question
682 views around the world
You can reuse this answer
Creative Commons License