Integral Calculus Question 376

Question: $ \int_{{}}^{{}}{{{(\log x)}^{2}}\ dx=} $

[IIT 1971, 77]

Options:

A) $ x{{(\log x)}^{2}}-2x\log x-2x+c $

B) $ x{{(\log x)}^{2}}-2x\log x-x+c $

C) $ x{{(\log x)}^{2}}-2x\log x+2x+c $

D) $ x{{(\log x)}^{2}}-2x\log x+x+c $

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Answer:

Correct Answer: C

Solution:

$ \int_{{}}^{{}}{{{(\log x)}^{2}}dx} $ . Put $ \log x=t\Rightarrow e^{t}=x\Rightarrow dx=e^{t}dt, $ then it reduces to $ \int_{{}}^{{}}{t^{2}.,e^{t}dt=t^{2}e^{t}-2te^{t}+2e^{t}+c} $
$ =x{{(\log x)}^{2}}-2x\log x+2x+c $ .