ln x dx x ln x - x c   Proof ln Consider ln x dx = x dx ∫ ∫ 1 dv Let u =ln x dx 1  dx Then 1 v = = x  ∫ 1dx x du Using the formula dv du 1 lnx dx  x ln x - dx x x = x ln x dx –∫ = x ln x x + c – Example 7 Find ∫ (ln x)2 dx Solution dx  (lnx)2.1dx 2 dv Let u = (lnx)2 dx 1  du 1 dx 2lnx  v= 1 dx = x ∫ x 40 Substitution of the results in the formula dv du dx x(lnx) - 2x(ln  2 2 1 dx x
ln x dx  x ln x - x  c Proof Consider ∫ ln x dx = ∫ ln x 1 dx Let u =ln x dv dx  1 Then du dx  1 v = ∫ 1dx = x x Using the formula dv du lnx dx  x ln x - x 1 x dx = x ln x –∫ dx = x ln x – x + c Example 7 Find ∫ (ln x)2 dx Solution 2 dx  (lnx)2.1dx Let u = (lnx)2 dv dx  1 du dx  2lnx 1 x v= ∫ 1 dx = x 40 Substitution of the results in the formula dv du 2 dx  x(lnx) 2 - 2x(ln 1 x dx