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Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 39

Answers (1)

Answer:

        log\left | x+log\, x \right |+C    

Hint:

Put

        x+log\, x=t

        (x+\frac{1}{x})dx=dt

        \frac{x+1}{x}dx=dt

Given:

        \int \! \frac {x+1}{x(x+log\, x)}dx                        ....(1)

Explanation:

Let

        x+log\, x=t

        (x+\frac{1}{x})dx=dt

        (\frac{x+1}{x})dx=dt

Put in (1) we get

        \int \! \frac{dt}{t}

        =log\left | t \right |+C

        =log\left | x+log\, x \right |+C

Posted by

Gurleen Kaur

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