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int (log (x+1) - log x)/(x(x-1)) dx is ...

`int (log (x+1) - log x)/(x(x-1)) dx ` is equal to :

A

`-1/2[In((x+1)/x)]^(2)+C`

B

`C-[{In(x+1)}^(2)-(Inx)^(2)]`

C

`-In[In((x+1)/x)]+C`

D

`-In((x+1)/x)+C`

Text Solution

Verified by Experts

The correct Answer is:
A
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int(log(x+1)-log x)/(x(x+1))dx= (A) log(x-1)log x+(1)/(2)(log x-1)^(2)-(1)/(2)(log x)^(2)+c (B) (1)/(2)(log(x+1))^(2)+(1)/(2)(log x)^(2)-log(x+1)log x+c (C) -(1)/(2)(log(x+1)^(2))-(1)/(2)(log x)^(2)+log x*log(x+1)+c (D) [log(1+(1)/(x))]^(2)+c