What is the integration of #int1/(1-t)dt - int 1/(2-t)dt# ?

#int1/(1-t)dt - int 1/(2-t)dt#

1 Answer

#\ln|\frac{t-2}{t-1}|+C#

Explanation:

Given that

#\int \frac{1}{1-t}\ dt-\int \frac{1}{2-t}\ dt#

#=-\int \frac{(-1)}{1-t}\ dt+\int \frac{(-1)}{2-t}\ dt#

#=-\int \frac{d(1-t)}{1-t}+\int \frac{d(2-t)}{2-t}#

#=-\ln|1-t|+\ln|2-t|+C#

#=\ln|\frac{2-t}{1-t}|+C#

#=\ln|\frac{t-2}{t-1}|+C#