Question

How do you find the integration of `log x`?

Answer 1

`int\ log(x)\ dx=1/ln(10)(xln(x)-x)+C=x/ln(10)(ln(x)-1)+C`

Explanation:

`int\ log(x)\ dx=int\ ln(x)/ln(10)\ dx`
`=1/ln(10)int\ ln(x)\ dx`
Using the integration by parts :
`int\ f(x)g'(x)\ dx=[f(x)g(x)]-int\ f'(x)g(x)\ dx`
There : `f(x)=ln(x), f'(x) =1/x,g(x)=x,g'(x)=1`
So: `int\ log(x)\ dx=1/ln(10)(xln(x)-int\ dx)`
So:`int\ log(x)\ dx=1/ln(10)(xln(x)-x)+C=x/ln(10)(ln(x)-1)+C`

In general, `int\ log_"n"(x)\ dx=x/ln(n) (ln(x)-1)+C`

`n in RR""_+^*` \ `{1}`, `C in RR`

Answer 2

`intlog_10(x)dx=[x(ln(x)-1)]/(ln(10))+C`

Explanation:

Remember that:

`log_a(b)=log_c(b)/(log_c(a))`

`=>log_10(x)=ln(x)/ln(10)`

We now have:

`intln(x)*1/(ln(10))dx`

`1/(ln(10))intln(x)dx`

Integration by parts:

`intudv=uv-intvdu`

We let:

`u=ln(x)`

`dv=1`

`=>du=d/dx(ln(x))`

`=>du=1/x`

`=>v=int1dx`

`=>v=x`

`=>1/(ln(10))*[xln(x)-intx*1/xdx]`

`=>1/(ln(10))*[xln(x)-int1dx]`

`=>1/(ln(10))*[xln(x)-x]`

`=>[xln(x)-x]/(ln(10))`

`=>[x(ln(x)-1)]/(ln(10))` Do you `C` why this is incomplete?

`=>intlog(x)dx=[x(ln(x)-1)]/(ln(10))+C`