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explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.26 question 12 maths

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Answer:
The correct answer is e^{x} \cdot \frac{1}{(1-x)}+c
Hint:

\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+c

Given:

\int \frac{2-x}{(1-x)^{2}} e^{x} d x

Solution:

        I=\int \frac{2-x}{(1-x)^{2}} e^{x} d x

            \begin{aligned} &=\int e^{x}\left(\frac{1+1-x}{(1-x)^{2}}\right) d x \\ &=\int e^{x}\left(\frac{1}{(1-x)^{2}}+\frac{1}{(1-x)}\right) d x \end{aligned}

we have,

\int e^{x}\left\{f(x) d x+f^{\prime}(x)\right\} d x=e^{x} f(x)+c

            \begin{aligned} &=\int e^{x}\left(\frac{1}{(1-x)^{2}}+\frac{1}{(1-x)}\right) d x \\ &=e^{x} \cdot \frac{1}{(1-x)}+c \end{aligned}

So, the correct answer is  e^{x} \cdot \frac{1}{(1-x)}+c

 

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