Need help integrating

[K.I.L.E.R]

S ln(x)/x dx

=

1 - (1/x)

=

x - ln(x)

?

 

[OpenGL guy]

S ln(x)/x dx

=

1 - (1/x)

=

x - ln(x)

?

Not exactly. First, ln(x)/x != 1 - 1/x. Second, the best way to solve this type of integral is by substitution. Notice the function ln(x) and it's derivative 1/x are present. This implies the substitution u = ln(x). Then du = 1/x dx. Thus your integral becomes S u du = 1/2 u^2 + C. Replace u with ln(x) to get final answer of 1/2 (ln(x))^2 + C.

 

[StefanS]

Or try partial integration:

§ lnx/x=(lnx)^2-§lnx/x
-> 2 § lnx/x=(lnx)^2
->§lnx/x=(lnx)^2+c

 

[K.I.L.E.R]

u = lnx
du = 1/x
v = 1/x
dv = lnx

u.v' = u.v - u'.v

lnx.lnx = lnx^2 - 1/x . lnx

= lnx^2 - lnx/x

Why is my answer so different?

 

[StefanS]

u = lnx
du = 1/x
v = 1/x
dv = lnx

u.v' = u.v - u'.v

lnx.lnx = lnx^2 - 1/x . lnx

= lnx^2 - lnx/x

Why is my answer so different?

Because partial integration is:

§ u*v'= u*v - § u' *v

you forgot the integral.

d/dx(u*v)=u'*v+u*v' -> u*v= §u'*v+§u*v' -> §u*v'=u*v-§u'*v

 

[OpenGL guy]

Or try partial integration:

§ lnx/x=(lnx)^2-§lnx/x
-> 2 § lnx/x=(lnx)^2
->§lnx/x=(lnx)^2+c

Your last step is incorrect. Since you are dividing both sides by two, the right hand side would be 1/2 (lnx)^2 + c. Also, you need a "+c" on the second line.

 

[StefanS]

Your last step is incorrect. Since you are dividing both sides by two, the right hand side would be 1/2 (lnx)^2 + c. Also, you need a "+c" on the second line.

Yeah, I forgot the 1/2. The +c is entirely irrelevant, though since it is an arbitrary constant to be fixed by the conditions imposed on the result. Call me lazy but i didn't want to redefine c_const/2=c.

 

[OpenGL guy]

Yeah, I forgot the 1/2. The +c is entirely irrelevant, though since it is an arbitrary constant to be fixed by the conditions imposed on the result. Call me lazy but i didn't want to redefine c_const/2=c.

But your second line states that twice an integral equals (lnx)^2 when it really should be (lnx)^2 + c. If you divide by two and still write c, that's fine as c is an arbitrary constant.

 

[StefanS]

But your second line states that twice an integral equals (lnx)^2 when it really should be (lnx)^2 + c. If you divide by two and still write c, that's fine as c is an arbitrary constant.

Oh boy, that's about as important as reflectors on your school bag. I am a physicist, we're all sloppy. Carrying arbitrary constants through all your calculations is a hassle, nobody does it. As long as you choose to add them in your end result, it is fine.