If `intx log (1+(1)/(x))dx`
`=f(x).log_(e)(x+1)+g(x)log_(e)x^(2)xLx+C` , then

Options

A. `f(x)=(x^(2))/(2)`
B. `g(x)=log_(e)x`
C. L = 1
D. `L=(1)/(2)`

Answer

d

Explanation

Let `I=intx log(1+(1)/(x))dx`
`rArrI=intunderset(II)(x)logunderset(I" ")((1+x))dx-intunderset(II)(x)logx dx`
`rArr I=(x^(2))/(2)log(1+x)-(1)/(2)int(x^(2))/(x+1)dx-{(x^(2))/(2)logx-int(1)/(x)xx(x^(2))/(2)dx}`
`rArr I=(x^(2))/(2)log_(e)(1+x)-(1)/(2)intx-1+(1)/(x+1)dx-(x^(2))/(2)log_(e)x+(x^(2))/(4)+C`
`rArrI=(x^(2))/(2)log_(e)(1+x)-(1)/(2)((x^(2))/(2)-x)-(1)/(2)log_(e)(x-1)-(x^(2))/(2)log_(e)x+(x^(2))/(4)+C`
`rArrI=((x^(2)-1)/(2))log_(e)(1+x)-(x^(2))/(2)log_(e)x+(x)/(2)+C`
Hence , f(x)`=(x^(2)-1)/(2) ,g(x)=-(x^(2))/(2)andL=(1)/(2)`