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provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.26 question 2

Answers (1)

Answer:
The correct answer is \frac{e^{x}}{x^{2}}+c
Hint: using integration by parts, \int u . v d x=u \int v d x-\int\left[\int v d x \frac{d u}{d x} d x\right]

Given:

Solution:

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

        =\int e^{x} \cdot \frac{1}{x^{2}} d x-2 \int \frac{e^{x}}{x^{3}} d x

        =\int e^{x} \cdot x^{-2} d x-2 \int \frac{e^{x}}{x^{8}} d x

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

        =\frac{e^{x}}{x^{2}}+c

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

 

 

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