Question

integration of 1/ log x dx

Answer 1

The derivative of ln(x) is 1/x. Integration goes the other way: the integral (or anti-derivative) of 1/x should be a function whose derivative is 1/x. As we just saw, this is ln(x). However, if x is negative then ln(x) is undefined! The solution is quite simple: the antiderivative of 1/x is ln(|x|).

so, the logarithmic integral is,

∫1log(x)dx=li(x)∫1log(x)dx=li(x)

which is known as the integral logarithm.

it can represent as

∫1log(x)dx=[math]∫1log(x)dx=[/math]

Let t=log(x)[math]t=log(x)[/math] which can also be written as et=x[math]et=x[/math]

Therefore on differentiating t=log(x)[math]t=log(x)[/math]on both sides we get,

dt=1xdx[math]dt=1xdx[/math]

or, dx=xdt→dx=etdt[math]dx=xdt→dx=etdt[/math]

∫1log(x)dx=∫ettdt[math]∫1log(x)dx=∫ettdt[/math]

which gives Ei(t)+c[math]Ei(t)+c[/math]

i.e Ei(log(x))+c[math]Ei(log(x))+c[/math]

Here Ei is an exponential integral.