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 $ .