Question

integrate 1/(x-1)^2 dx​

Answer 1

Solution :

$:\implies \sf y = \displaystyle \int \sf \dfrac{1}{(x - 1)^{2}}dx$

$\textsf{Now, by applying the Substitution method of integration and putting (x - 1) as u, we get :-}$

$\textsf{By differentiating (x - 1), we get :} \\ \\ \textsf{Function of u = (x - 1)} \\ \\ :\implies \sf{\dfrac{du}{dx} = \dfrac{d(x - 1)}{dx}} \\ \\ :\implies \sf{\dfrac{du}{dx} = \dfrac{d(x)}{dx} - \dfrac{d(1)}{dx}} \\ \\ :\implies \sf{\dfrac{du}{dx} = 1 - 0} \\ \\ :\implies \sf{\dfrac{du}{dx} = 1} \\ \\ \therefore \sf{du = dx} \\ \\ \textsf{Hence, we get that du = dx}$

$\textsf{Now, by substituting the above values in the equation, we get :-}$

$:\implies \sf y = \displaystyle \int \sf \dfrac{1}{u^{2}}du$

$:\implies \sf y = \displaystyle \int \sf u^{-2}du$

$\textsf{Now, by applying the power rule of integration, we get :} \\ \\ \underline{\sf{\bigstar\: Power\:rule\:of\: integration :}} \\ \\ :\implies \displaystyle \int \sf x^{n} = \dfrac{x^{(n + 1)}}{n + 1} + C\: Where, n \ne -1$

$:\implies \sf y = \displaystyle \int \sf u^{-2}du = \dfrac{u^{(-2 + 1)}}{-2 + 1} + C$

$:\implies \sf y = \displaystyle \int \sf u^{-2}du = \dfrac{u^{-1}}{-1} + C$

$:\implies \sf y = \displaystyle \int \sf u^{-2}du = -\dfrac{1}{u} + C$

$\textsf{Now, by substituting the value of u (i.e, (x - 1) in the equation, we get :-}$

$:\implies \sf y = \displaystyle \int \sf u^{-2}du = -\dfrac{1}{(x - 1)} + C$

$\underline{\therefore \sf y = \displaystyle \int \sf u^{-2}du = -\dfrac{1}{(x - 1)} + C}$