1/x(lnx)^2?

Question

How to integrate 1/x(lnx)^2 dx ?
i just take lnx=y
so,dy=dx/x.
but after that i can't do. please solve this problem.

Answer 1

yor are almost close to the solution
so the given expression

1/x(ln x)^2 dx becomes 1/y^2 dy oy dy/y^2

integrating this we get -1/y or -1/ln x

Answer 2

You're right so far, now solve that equation for dx. You get dx=x dy. Substitute, and you get ∫1/(xy²) * x dy. Cancel the x, and you get ∫1/y² dy, which is -1/y. Then simply resubstitute ln x to get your final answer of -1/ln x.

P.S. your initial problem should have been written 1/(x (ln x)²) - the way you wrote it might be misinterpreted as being equivalent to (ln x)²/x.

Answer 3

Once you have dy/y_square, apply the first case of integrals, which is x_n --> nx_n-1 but inversely.

Then restitute y. The answer is -1/lnx.

Answer 4

dx / [x*(ln(x))^2] = dx * (1/x) * (ln(x))^-2 = dx * (1/x) * (ln(x))^-1 *(ln(x))^-1

U-Substitution:

u = (ln(x))^-1 and du = -(1/x)(ln (x))^-2 dx

Integral {dx / [x*(lnx)^2]} = Integral {-du}

= -u + C
=(ln(x))^-1 + C
or 1/[ln(x)] + C