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If xlog(x+1x)dx=f(x)log(x+1)+g(x)x2+Lx+C then

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a
f(x)=(1/2)x2−1
b
g(x)=log⁡x
c
L = 1
d
none of these

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detailed solution

Correct option is A

The given integral equals ∫xlog⁡x+1xdx=∫xlog⁡(x+1)dx−∫xlog⁡xdx=x22log⁡(x+1)−12∫x2x+1dx−x22log⁡x+12∫x2xdx=x22log⁡(x+1)−x22log⁡x−12∫x−1+1x+1dx+12∫xdx=x22log⁡(x+1)−x22log⁡x+x2−12log⁡(x+1)+C⇒f(x)=x22−12,g(x)=−12log⁡x and L=12.


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