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.

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

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

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

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

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

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