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  • need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.26 question 23

need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.26 question 23

Answers (1)

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Answer:
The correct answer is  e^{x} \frac{1}{(x-2)^{2}}+c
Hint:

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

Given:

\int e^{x} \frac{(x-4)}{(x-2)^{3}} d x

Solution:

        \begin{aligned} &I=\int e^{x} \frac{(x-4)}{(x-2)^{3}} d x \\ &=\int e^{x} \frac{(x-2-2)}{(x-2)^{3}} d x \\ &=\int e^{x}\left[\frac{(x-2)}{(x-2)^{3}}-\frac{2}{(x-2)^{3}}\right] d x \end{aligned}

        =\int e^{x}\left[\frac{1}{(x-2)^{2}}+\frac{-2}{(x-2)^{3}}\right] d x

Now, consider,

        f(x)=\frac{1}{(x-2)^{2}}

Then  f^{\prime}(x)=\frac{-2}{(x-2)^{3}}

Thus the above integral is of the form

        e^{x}\left(f(x)+f^{\prime}(x)\right)

As,

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

        I=e^{x} \frac{1}{(x-2)^{2}}+c

So, the correct answer is  e^{x} \frac{1}{(x-2)^{2}}+c

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