Question

integration of log x​

Answer 1

Answer

∫log x dx = x log x -x

Proof : Using integration by parts,

∫udv = uv - ∫vdu

In ∫ log x dx,

take, u=logx => du= (1/x) . dx

∫dv=∫dx => v=x

Now substituting,

∫log x = logx (x) - ∫x . 1/x . dx = logx (x) - ∫dx= x log x -x + C

where C is constant.

Hope it helps!!