Question

Integrate `lnx/10^x`?

Answer 1

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`

Answer 2

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.

Answer 3

`(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