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The primitive of \log x will be

  • Option 1)

    x\log (e+x) +c

  • Option 2)

    x\log \frac{e}{x} +c

  • Option 3)

    x\log \frac{x}{e} +c

  • Option 4)

    x\log ex +c

 

Answers (1)

As learnt in

Integration By PARTS -

Let u and v are two functions then 

\int u\cdot vdx=u\int vdx-\int \left ( \frac{du}{dx}\int vdx \right )dx

- wherein

Where u is the Ist function v is he IInd function

 

 

\int \log x d x= \int 1. \log xdx \ \ \ \ \\

 

                       = \int 1. \log x d x -\int (\frac{d}{dx} (\log x)\int 1.dx) dx

                       = x \log x \int \frac{1}{x}. xdx

                       = x \log x-x =x log\frac{x}{e}+c   


Option 1)

x\log (e+x) +c

This option is incorrect.

Option 2)

x\log \frac{e}{x} +c

This option is incorrect.

Option 3)

x\log \frac{x}{e} +c

This option is correct.

Option 4)

x\log ex +c

This option is incorrect.

Posted by

Vakul

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