Integrate lnx/10^x?

3 Answers

mistake

Explanation:

int(lnx)/10^xdx can also be written as int(lnx)xx10^(-x)dx.

Now, we can use the formula for integral of product

intu*v*dx=u*v-int(v*du), where u=lnx

As such, we have du=(1/x)dx and let dv=x^(-10)dx or v=x^(-9)/-9

Hence, intu*v*dx=(-1/9)lnx.x^(-9)-int(x^(-9)/-9)*dx/x, or

= (-1/9)lnx.x^(-9)+(1/9)intx^(-10)*dx

= (-1/9)lnx.x^(-9)+(1/9)x^(-9)/(-9)+c

= (-1/9)lnx.x^(-9)-(1/81)x^(-9)+c

= -1/81(x^(-9))(9lnx+1)+c

Mar 12, 2016

Appears infinite series integral to me.

Explanation:

We can use the formula for integral of product of two function u(x) and v(x)

intucdotdv=ucdotv-int vcdotdu

(rule can be simply be derived by integrating the product rule of differentiation)
Given integral intln(x)//10^xcdotdx can be written as
intln(x)xx10^(-x)cdotdx

Let u=ln(x) and dv=10^(-x)cdot dx
from first assumption du=1/x cdotdx
from the second equality v=int 10^-x cdot dx=-1/ln 10 10^-x+C

We get intln(x)xx10^(-x)cdotdx=ln(x)cdot(-1/ln 10 10^-x+C)-int(-1/ln 10 10^-x+C)cdot 1/xcdot dx
Where C is a constant of integration.
=ln(x)cdot(-1/ln 10 10^-x+C)+int1/ln 10 10^-xcdot 1/xcdot dx-intCcdot 1/xcdot dx
=ln(x)cdot(-1/ln 10 10^-x+C)+int1/ln 10 10^-xcdot 1/xcdot dx-Ccdot ln|x|+C_2,simplifying
=ln(x)cdot(-1/ln 10 10^-x)+1/ln 10 int 10^-xcdot 1/xcdot dx+C_2

It reduces to finding the integral of intx^-1cdot 10^-xcdot dx
Again using the above integral by parts formula
Let u=x^-1 and dv=10^(-x)cdot dx
du=-x^-2cdot dx and we already have the value for v
intx^-1cdot 10^-xcdot dx=x^-1cdot (-1/ln 10 10^-x+C)-int(-1/ln 10 10^-x+C)cdot (-x^-2cdot dx)

  1. Inspection reveals it turns out to be finding int 10^-xcdot x^-2cdot dx and so on.
  2. Function ln (x) is defined only for x>0
  3. The integral appears to be infinite series integral.
Mar 15, 2016

(lny)(ln(ln_10 y))-lny = (lny)(ln(ln_10 y)-1)

Then put in 10^x for y

(ln 10^x )(ln(ln_10 10^x)-ln 10^x

Explanation:

Let y=10^x

lny=ln10^x

lny=x*ln10

x=lny/ln10 = ln_10y=log_10exxlog_e y
:.dx=log_10exx1/yxxdy

int (ln(ln_10 y))/yxxlog_10exx1/yxxdy
=int (ln(ln_10 y))/y^2xxlog_10exxdy ; u=ln(ln_10 y)=ln(1/ln10 *lny), dv=1/y

du=1/(ln y/ln10) *1/(yln10 ) = (ln10/lny)(1/(yln10))=1/(ylny)

v=lny

uv-intvdu ->( ln(ln_10 y))lny-intlny *1/(ylny)

(lny)(ln(ln_10 y))-int1/y

(lny)(ln(ln_10 y))-lny = (lny)(ln_10 y-1)

Then put in 10^x for y

ln 10^x (ln(ln_10 10^x)-ln 10^x

PROOF:
d/dy((lny)(ln(ln_10 y) -1))

f=lny, g=ln(ln_10 y) -1)

f'=1/y, g'=(1/ln_10y)(1/(yln10))

fg'+gf'---> product rule
lny *(1/ln_10y)(1/(yln10))+(ln(ln_10y) -1) * 1/y

lny (1/(lny/ln10))(1/(yln10))+(ln(ln_10y) -1) * 1/y

lny (ln10/lny)(1/(yln10))+(ln(ln_10y) -1) * 1/y

1/y +(ln(ln_10 y) -1)/y

((1 +ln(ln_10 y) -1))/y

(ln(ln_10y))/y

ln(x)/10^x--->ln_10 y =x from above