Question #57356

2 Answers
Monzur R.
May 26, 2017

#int (x-1)/(x^2-x)# #dx=int(x-1)/(x(x-1)) # #dx=int1/x # #dx=ln|x|+"c"#

Nityananda
May 26, 2017

log x +c

Explanation:

Given, #int [x-1]/[x^2-x] dx#

#rArr int [x-1]/[x(x-1)]dx#

#rArr int cancel[(x-1)]/[x (cancel[x-1])]dx#

#rArr int 1/x dx#

#rArr logx + c#

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