How do you evaluate the integral #int (2xdx)/(x-1)#?

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Answer 1 · 2017-12-19T14:46:02

The integral equals #2x + 2ln|x- 1| + C#

I would use partial fractions.

#A/1 + B/(x- 1) = (2x)/(x- 1)#

#A(x -1) + B = 2x#

#Ax - A + B = 2x#

#Ax + (B - A) = 2x#

From here it's clear the #A = 2# and #B - A = 0# therefore, #B = 2#.

#I = int 2 + 2/(x -1)dx#

#I = 2x + 2ln|x- 1| + C#

Hopefully this helps!

Answer 2 · 2017-12-19T17:27:26

Another way to see the rewrite.

#int (2xdx)/(x-1)dx = 2 int x/(x-1) dx#

# = 2int ((x-1)+1)/(x-1) dx#

# = 2int ((x-1)/(x-1)+1/(x-1)) dx#

# = 2int (1+1/(x-1)) dx#

# = 2(x+lnabs(x-1))+C#