ddxlog|x|=….(x≠0)
1x
-1x
x
-x
Finding the value:
Given, y=log|x|….(x≠0)
As we know,
x=xx≥0-xx<0
But log is not defined for x<0
⇒logx=logx
Differentiate it with respect to ‘x’, we get
∴d(log|x|)dx=1x
Hence, option (A) is correct.
∫a0f(x)dx=∫a0f(a−x)dx and ∫2a0f(x)dx=2∫a0f(x)dx If f(2a−x)=f(x)