Question

Intrigation of 1/X(1 logx)DX

Answer 1

Answer:= log |1 + logx| + c .

Step-by-step explanation:

If (1+log x ) is in the numerator,then

i.e § (1+logx )/x dx

Splitting into two fractions we get

§ 1/x dx + § logx /x dx

[ Taking logx = z ;

dz = 1/x dx ]

= log x + § z dz

= log x + z^2 /2

= log x + log x^2 /2

= logx [ 1 + (log x)/2] + c .

If (1+logx) is in the denominator.

[Taking 1 +logx =z

dz = 1/x dx]

§ dx / x (1+logx)

= § dz /z

= log z + c

= log |1 + logx| + c .