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  • explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.26 question 16 maths

explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.26 question 16 maths

Answers (1)

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

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

Given:

\int e^{x}\left(\log x+\frac{1}{x^{2}}\right) d x

Solution:

        I=\int e^{x}\left(\log x+\frac{1}{x^{2}}\right) d x

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

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

We have,

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

            =e^{x}(\log x)+e^{x}\left(\frac{-1}{x}\right)+c

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

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