The value of ∫x2+1x2-1·dx is
log(x–1)x+1+c
log(x+1)x-1+c
x+log(x–1)x+1+c
log(x-1)2+c
Explanation for the correct option:
Compute the required value,
Given: ∫x2+1x2-1·dx
∫x2-1+2x2-1·dx=∫x2-1x2-1·dx+∫2x2-1·dx=∫1·dx+∫2x2-1·dx=x+c+∫x+1-(x-1)x2-1·dx=x+c+∫x+1-(x-1)(x-1)(x+1)·dx=x+c+∫1(x-1)·dx-∫1(x+1)·dx=x+c1+log(x-1)+c2-log(x+1)+c3∫1x·dx=log(x)=x+logx-1x+1+C
where c1,c2,c3,C are constants
Hence, option C is the correct answer.