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

2 Answers

Answer:

#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

Answer:

#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#