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Evaluate the indefinite integral:
[tex]\mathsf{\displaystyle\int\!\dfrac{1}{(1-y)^2}\,dy}\\\\\\ =\mathsf{\displaystyle\int\!\dfrac{-1}{(1-y)^2}\cdot (-1)\,dy}\\\\\\ =\mathsf{\displaystyle- \int\!(1-y)^{-2}\cdot (-1)\,dy\qquad\quad(i)}[/tex]
Substitution:
[tex]\begin{array}{lcl} \mathsf{1-y=u}&\quad\Rightarrow\quad&\mathsf{y=1-u}\\\\ &&\mathsf{dy=-1\,du} \end{array}[/tex]
So the integral [tex]\mathsf{(i)}[/tex] becomes
[tex]=\mathsf{\displaystyle -\int\! u^{-2}\,du}\\\\\\ =\mathsf{-\,\dfrac{u^{-2+1}}{-2+1}+C}\\\\\\ =\mathsf{-\,\dfrac{u^{-1}}{-1}+C}\\\\\\ =\mathsf{u^{-1}+C}\\\\\\ =\mathsf{\dfrac{1}{u}+C}[/tex]
[tex]=\mathsf{\dfrac{1}{1-y}+C}\\\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{\displaystyle\int\!\dfrac{1}{(1-y)^2}\,dy=\dfrac{1}{1-y}+C} \end{array}}\qquad\quad\checkmark[/tex]
I hope this helps. =)
Tags: indefinite integral fraction rational substitution power differential calculus
