1.∫dx/logxwhat will value of the intigarlRgcb .oith lgfh boryi kgyjc gegv

Well, the logarithmic integral is given by -

∫1log(x)dx=li(x)∫1log(x)dx=li(x)

which is a special function known as the integral logarithm.

It can also be represented as -

∫1log(x)dx=∫1log(x)dx=

Lett=log(x)which can also be written ase^t=x
Therefore on differentiatingt=log(x)on both sides we get,

dt=xdx

or,dx=xdt→dx=etdt

∫1log(x)dx=∫e^t/tdt

which givesEi(t)+c

i.eEi(log(x))+c

HereEi is an exponential integral.

Well, the logarithmic integral is given by -

∫1log(x)dx=li(x)∫1log(x)dx=li(x)

which is a special function known as the integral logarithm.

It can also be represented as -

∫1log(x)dx=∫1log(x)dx=

Lett=log(x)which can also be written ase^t=x
Therefore on differentiatingt=log(x)on both sides we get,

dt=xdx

or,dx=xdt→dx=etdt

∫1log(x)dx=∫e^t/tdt

which givesEi(t)+c

i.eEi(log(x))+c

HereEi is an exponential integral.

Approved