Question

integrate e^x(2-x /(1-x)^2)dx
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Answer 1

We know,

$1. \: \boxed{ \pink{ \rm \: {e}^{x} (f(x) \: + \: f'(x)) \: = \: {e}^{x} \: f(x) \: + \: c}}$

Consider,

$\rm : \implies \: Let \: I \: = \int \: {e}^{x}\dfrac{2 - x}{ {(1 - x)}^{2} } dx$

$\rm : \implies \: \int \: {e}^{x}\dfrac{1 + 1 - x}{ {(1 - x)}^{2} } dx$

$\rm : \implies \: \int \: {e}^{x}\dfrac{1 + (1 - x)}{ {(1 - x)}^{2} } dx$

$\rm : \implies \: \int \: {e}^{x} \bigg(\dfrac{1}{ {(1 - x)}^{2} } + \dfrac{(1 - x)}{ {(1 - x)}^{2} } \: \bigg)$

$\rm : \implies \: \int \: {e}^{x} \bigg(\dfrac{1}{ {(1 - x)}^{2} } + \dfrac{1 }{ {(1 - x)}} \: \bigg)$

$\bf \: Let \: f(x) \: = \: \dfrac{1}{1 - x}$

On differentiating w. r. t. x both sides, we get

$\rm : \implies \: f'(x) \: = \dfrac{d}{dx} {(1 - x)}^{ - 1}$

$\rm : \implies \: f'(x) \: = \: - 1 \: {(1 - x)}^{ - 2} \dfrac{d}{dx} (1 - x)$

$\rm : \implies \: f'(x) \: = \: - 1 \: \times {(1 - x)}^{ - 2} ( - 1)$

$\rm : \implies \: f'(x) \: = \: \: {(1 - x)}^{ - 2}$

$\rm : \implies \: f'(x) \: = \: \dfrac{1}{ {(1 - x)}^{2} }$

Hence,

$\rm : \implies \: \int \: {e}^{x} \bigg(\dfrac{1}{ {(1 - x)}^{2} } + \dfrac{1 }{ {(1 - x)}} \: \bigg) = {e}^{x} \dfrac{1}{1 - x} + c$