`Ifintxlog(1+1/x)dx=f(x)log(x+1)+g(x)x^2+A x+C ,` then `f(x)=1/2x^2` (b) `g(x)=logx` `A=1` (d) none of these

Options

A. `f(x)=(1)/(2)x^(2)`
B. `g(x)=log x`
C. `A=1`
D. none of these

Answer

D

Explanation

` intx log(1+(1)/(x))dx `
`=int xlog(x+1)dx-int x logxdx `
` =(x^(2))/(2)log(x+1)-(1)/(2)int(x^(2))/(x+1)dx-(x^(2))/(2)logx+(1)/(2)int(x^(2))/(x)dx+C`
` =(x^(2))/(2)log(x+1)-(1)/(2)int(x-1+(1)/(x+1))dx-(x^(2))/(2)logx+(1)/(4)x^(2)+C`
`=(x^(2))/(2)log(x+1)-(x^(2))/(2)logx-(1)/(2)((x^(2))/(2)-x)-(1)/(2)log(x+1)+(1)/(4)x^(2)+C`
`=(x^(2))/(2)log(x+1)-(x^(2))/(2)logx-(1)/(2)log(x+1)+(1)/(2)x+C`
`"Hence, "f(x)=(x^(2))/(2)-(1)/(2),g(x)=-(1)/(2)logx, " and " A=(1)/(2).`