How do you evaluate the integral∫(x-1)(x+1)dx ?
Step 1: Split the integrating function:
∫(x-1)(x+1)dx
=∫x+1-2(x+1)dx=∫x+1x+1-2x+1dx=∫1-2x+1dx
Step 2: Integrate the function using the linearity property.
∫1-2x+1dx=∫dx-∫2x+1dx=x-2∫1x+1dx....................(1)
Let x+1=u
⇒ dx=du
Thus,
∫1x+1dx=∫1udu=lnu=lnx+1 ∫1xdx=lnx
Substitute the value in equation 1,
∫1-2x+1dx=x-2lnx+1+c
Hence, the value of ∫(x-1)(x+1)dx is x-2lnx+1+c.
How do you evaluate log1?