Integrate lnx/10^xlnx10x?

3 Answers

mistake

Explanation:

int(lnx)/10^xdxlnx10xdx can also be written as int(lnx)xx10^(-x)dx(lnx)×10xdx.

Now, we can use the formula for integral of product

intu*v*dx=u*v-int(v*du)uvdx=uv(vdu), where u=lnxu=lnx

As such, we have du=(1/x)dxdu=(1x)dx and let dv=x^(-10)dxdv=x10dx or v=x^(-9)/-9v=x99

Hence, intu*v*dx=(-1/9)lnx.x^(-9)-int(x^(-9)/-9)*dx/xuvdx=(19)lnx.x9(x99)dxx, or

= (-1/9)lnx.x^(-9)+(1/9)intx^(-10)*dx(19)lnx.x9+(19)x10dx

= (-1/9)lnx.x^(-9)+(1/9)x^(-9)/(-9)+c(19)lnx.x9+(19)x99+c

= (-1/9)lnx.x^(-9)-(1/81)x^(-9)+c(19)lnx.x9(181)x9+c

= -1/81(x^(-9))(9lnx+1)+c181(x9)(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)u(x)andv(x)

intucdotdv=ucdotv-int vcdotduudv=uvvdu

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

Let u=ln(x) and dv=10^(-x)cdot dxu=ln(x)anddv=10xdx
from first assumption du=1/x cdotdxdu=1xdx
from the second equality v=int 10^-x cdot dx=-1/ln 10 10^-x+Cv=10xdx=1ln1010x+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 dxln(x)×10xdx=ln(x)(1ln1010x+C)(1ln1010x+C)1xdx
Where CC 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)(1ln1010x+C)+1ln1010x1xdxC1xdx
=ln(x)cdot(-1/ln 10 10^-x+C)+int1/ln 10 10^-xcdot 1/xcdot dx-Ccdot ln|x|+C_2,=ln(x)(1ln1010x+C)+1ln1010x1xdxCln|x|+C2,simplifying
=ln(x)cdot(-1/ln 10 10^-x)+1/ln 10 int 10^-xcdot 1/xcdot dx+C_2=ln(x)(1ln1010x)+1ln1010x1xdx+C2

It reduces to finding the integral of intx^-1cdot 10^-xcdot dxx110xdx
Again using the above integral by parts formula
Let u=x^-1u=x1 and dv=10^(-x)cdot dxdv=10xdx
du=-x^-2cdot dxdu=x2dx and we already have the value for vv
intx^-1cdot 10^-xcdot dx=x^-1cdot (-1/ln 10 10^-x+C)-int(-1/ln 10 10^-x+C)cdot (-x^-2cdot dx)x110xdx=x1(1ln1010x+C)(1ln1010x+C)(x2dx)

  1. Inspection reveals it turns out to be finding int 10^-xcdot x^-2cdot dx10xx2dx and so on.
  2. Function ln (x)ln(x) is defined only for x>0x>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)(lny)(ln(ln10y))lny=(lny)(ln(ln10y)1)

Then put in 10^x10x for y y

(ln 10^x )(ln(ln_10 10^x)-ln 10^x(ln10x)(ln(ln1010x)ln10x

Explanation:

Let y=10^xy=10x

lny=ln10^xlny=ln10x

lny=x*ln10lny=xln10

x=lny/ln10 = ln_10y=log_10exxlog_e yx=lnyln10=ln10y=log10e×logey
:.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