integrate wrt x: (1/x)*logx?

(1/x)*logx*dx

= logx d(logx)

on integration

=1/2 (logx)^2 + C

you can write it as integral of logx*(1/x)dx. I am assuming its a natural log.

now as u can see derivative of logx (i.e 1/x) is invloved in integral thus u can apply power rule of integration.

solution is

integral of logx*1/x*dx ------(A)

substitue t=logx

now dt=1/x*dx

now statement A will look like this

integral of t*dt

on integrating you will get t^2 + C

now putting back the value of 't' will yield you

((logx)^2)/2 +c

n tht's ur answer!!!!

Substitute logx=t. diffrentiating, we get (1/x)dx=dt. Therefore the integrand becomes t.dt, whose integral is (t^2)/2 + c. Back substituting t as logx, we have the solution as (logx)^2/2 +c.

the given qty is logx (1/x)dx

= logx d(logx)

Integral = (1/2)*(logx)^2

Integral ( (1/x)log[base 10](x) )dx

{Note: By log(x), I'm going to assume base 10, the same way a calculator does. I'm NOT going to assume log[base e], because there's already an assigned function for that; ln}

I'm going to change the form of that slightly, so I can show you how the substitution is done.

Integral (log(x) (1/x)dx)

Before we perform our substitution, remember that the derivative of log[base a](x) is 1/[xln(a)]. It looks similar to the derivative of

ln(x) except it's multiplied by 1/ln(a).

Let u = log(x). Then

du = 1/[x ln(10)] dx

Multiplying both sides by ln(10), we have

ln(10) du = (1/x) dx

Note that we can now directly replace (1/x) dx with ln(10) du. The result of our substitution is then:

Integral (log(x) (1/x)dx) becomes

Integral (u (ln 10) du)

Let's pull out the constant ln 10 out of the integral.

(ln10) * Integral (u du)

And now, it's straightforward reverse power rule.

(ln10) * (1/2)u^2 + C

Substituting in u = log(x) back, we have

(ln10) * (1/2) (log(x))^2 + C

Simplifying this slightly,

[(ln10)/2] * [log(x)]^2 + C

∫(1/x)*logxdx

let u = logx

du = dx/x

∫(1/x)*logxdx = udu = (1/2)u^2 + C

∫(1/x)*logxdx = (1/2)(logx)^2 + C

First log x = ln x /ln10 , so wrt x = 1/ln10 *(1/x) *lnx

if you put u = lnx i/x *dx =du

so you must integer (1/ln10 ) udu

integration of udu is u^2/2

and the result of the integration is (1/2*ln10) (lnx)^2

put logx=t

1/x dx=dt;

it simply reduces to integral tdt which is t^2/2 +c

The answer is (logx)^2/2+c

assume In(x)=t, we have d(Inx)/dx=1/x so 1/x * dx =d(Inx)

so Integrate (1/x)* logx .wrt.x = Integ( 1/x) * Inx/ In(10) .wrt.x

as Integ (1/x) * Inx .wrt.x=integ Inx wrt.(Inx)

so Integ (1/x) * inx. wrt.x = integ t wrt. t = 0.5 t * t

as In (x) = t so

ur question = 0.5 (in(x) * In(x))/ In(10)

Integrate (1/x)(ln x). I assume you mean the natural log.

∫(1/x)(ln x)dx

Let

u = ln x

du = (1/x)dx

∫(1/x)(ln x)dx = ∫u(du) = u²/2 + C = (ln x)²/2 + C