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`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 theseA. `f(x)=(1)/(2)x^(2)`B. `g(x)=log x`C. `A=1`D. none of these |
Answer» Correct Answer - D ` 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).` |
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