Answer:
[tex]\large \boxed{\sf \ \ \ 2x-ln(1-x) \ \ \ }[/tex]
Step-by-step explanation:
Hello,
for any real x different from 1
[tex]\dfrac{3-2x}{1-x}=\dfrac{2-2x+1}{1-x}\\\\=\dfrac{2(1-x)+1}{1-x}\\\\=2\dfrac{1-x}{1-x}+\dfrac{1}{1-x}\\\\=2+\dfrac{1}{1-x}[/tex]
if f(x)=2x
f'(x)=2
if g(x)=-ln(1-x)
[tex]g'(x)=\dfrac{-1*-1}{1-x}=\dfrac{1}{1-x}[/tex]
So antiderivative of
[tex]\dfrac{3-2x}{1-x}=2+\dfrac{1}{1-x}[/tex]
Is
2x-ln(1-x)
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
