How do you find the integration of log x?

2 Answers

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

Apr 12, 2018

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