integral of logx/(1+logx)²dx
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integral = logx / (1+logx)^2 dx
Let log x = t
differentiating we get
d/dx logx = dt
1/x dx = dt
dt/dx = 1/x
SO
Integral = t / (1+t)^2 x dt
integrating we get
integtral = 2[ logx ] +c
Let log x = t
differentiating we get
d/dx logx = dt
1/x dx = dt
dt/dx = 1/x
SO
Integral = t / (1+t)^2 x dt
integrating we get
integtral = 2[ logx ] +c
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