Ques. -1

Calculus Level 5

∫ x log ⁡ ( 1 + 1 x ) d x = f ( x ) log ⁡ ( x + 1 ) + g ( x ) x 2 + x L + c \displaystyle \int x \log\left( 1+\frac{1}{x} \right) \, dx=f(x) \log(x+1)+g(x) x^2 +x L +c

Given the above integral with arbitrary constant c c , choose the right option.

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g(x)=logx None of the given options is true. f ( x ) = x 2 2 f(x)=\frac{x^2}{2} L=1

Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Bhargav Upadhyay
Mar 1, 2015

∫ x log ⁡ ( 1 + 1 x ) d x = ∫ x { log ⁡ ( x + 1 ) − log ⁡ x } = log ⁡ ( x + 1 ) ( x 2 2 − 1 2 ) + x 2 ( log ⁡ x 2 ) + x 2 + c ∴ f ( x ) = ( x 2 2 − 1 2 ) ∴ g ( x ) = ( x 2 2 − 1 2 ) ∴ L = 1 2 ∴ N o n o f t h e g i v e n o p t i o n i s t r u e . \int { x\quad \log { (1+\frac { 1 }{ x } ) } } dx\\ =\quad \int { x\quad \{ \log { (x+1) } -\log { x } \} } \\ =\log { (x+1) } (\frac { { x }^{ 2 } }{ 2 } -\frac { 1 }{ 2 } )+{ x }^{ 2 }\quad (\frac { \log { x } }{ 2 } )+\frac { x }{ 2 } +c\\ \therefore \quad f(x)=(\frac { { x }^{ 2 } }{ 2 } -\frac { 1 }{ 2 } )\\ \therefore g(x)=(\frac { { x }^{ 2 } }{ 2 } -\frac { 1 }{ 2 } )\\ \therefore L\quad =\quad \frac { 1 }{ 2 } \\ \therefore \quad Non\quad of\quad the\quad given\quad option\quad is\quad true.

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fx is x^2/2 wrong solution

Arsalan Abbas - 3 years, 2 months ago

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