Why 1 / (x + 1) DX equals ln (1 + x) Integral 1 / (x + 1) DX equals ln (1 + x)

Why 1 / (x + 1) DX equals ln (1 + x) Integral 1 / (x + 1) DX equals ln (1 + x)

∫1/(x+1) dx = ln(x+1) +C
∵[ln(x+1) +C]' = 1/(x+1)
∴∫1/(x+1) dx = ln(x+1) +C
C is a constant

The detailed process of finding ∫ ln (x / 3) DX

Using partial integration method
∫ln(x/3) dx
=ln(x/3) *x - ∫ x * d ln(x/3)
=ln(x/3) *x - ∫ x * 3/x *1/3 dx
=ln(x/3) *x - ∫dx
=Ln (x / 3) * x - x + C, C is a constant

∫ ln (x ^ 2 - 1) DX step

ln(x^2 -1)=ln(x+1)+ln(x-1)
∫ ln(x^2 -1)dx =∫ln(x+1)d(x+1)+∫ln(x-1)d(x-1)
Part points:
The original formula = (x + 1) ln (x + 1) -∫ (x + 1) d (LN (x + 1)) + (x-1) ln (x-1) -∫ (x-1) d (LN (x-1)) = (x + 1) ln (x + 1) -∫ (x + 1) * 1 / (x + 1) DX + (x-1) ln (x-1) -∫ (x-1) * 1 / (x-1) DX = (x + 1) ln (x + 1) + (x-1) ln (x-1)