Wolfram's wrong? O.o (1 Viewer)

[NubMuncher]

Integrate 1/[x(x^2 + 1)]

Wolfram says ln|x| - 1/2 ln |x^2 + 1| + C

I get ln|x| - tan^-1 x + C

 

[HyperComplexxx]

is

 

[Drongoski]

Integrate 1/[x(x^2 + 1)]

Wolfram says ln|x| - 1/2 ln |x^2 + 1| + C

I get ln|x| - tan^-1 x + C

I think you should get:

 

[NubMuncher]

I beg to differ, within our school we have not learned how to differentiate inverse trig functions yet.

Just realised that while doing partial fractions i forgot to add the x lmao, kinda fail XD

thanks

 

[laplace5]

I think you should get:

lol
why not just leave it as ln [ ( x) / ( sqrt(x^2+1) ) ] +C

or better yet let C= ln(A)

so answer = ln [ Ax / ( sqrt(x^2 +1) )] , where A is a positive constant

I love how logs simplify so much

or you do ln [ A sqrt [ (x^2) / (x^2+1) ]

nice and tidy

 

[jyu]

lol
why not just leave it as ln [ ( x) / ( sqrt(x^2+1) ) ] +C

or better yet let C= ln(A)

so answer = ln [ Ax / ( sqrt(x^2 +1) )] , where A is a positive constant

I love how logs simplify so much

or you do ln [ A sqrt [ (x^2) / (x^2+1) ]

nice and tidy

ln [ Ax / ( sqrt(x^2 +1) )] not= ln [ A sqrt [ (x^2) / (x^2+1) ]]

 

[laplace7]

no noob, they are both equivalent.

you get the second result if you use drongoskis answer, and express C as ln(A) and then use logA + logB = logAB

 

[jyu]

no noob, they are both equivalent.

you get the second result if you use drongoskis answer, and express C as ln(A) and then use logA + logB = logAB

they are not equivalent, hope you know why

 

[laplace7]

they are not equivalent, hope you know why

even using basic algebra they are equivalent

sqrt[ x^2 / (x^2 +1) ] = sqrt (x^2) / sqrt (x^2+1) = x/ sqrt(x^2+1) , yes?

or you are prob bullshitting about all the absolute values and shit in the log , lol, because what if x<0 , blah blah...



t

 

[jyu]

sqrt[ x^2 / (x^2 +1) ] = sqrt (x^2) / sqrt (x^2+1) = x/ sqrt(x^2+1) , yes?

or you are prob bullshitting about all the absolute values and shit in the log , lol, because what if x<0 , blah blah...



t

you are right, they cannot be equivalent if they have different domains

 

[jyu]

to illustrate the point further, sketch both functions and see they are the same

 

[jyu]

hope you know the meaning of 'equivalent'

 

[RANK 1]

wolfram's never wrong, i sometimes think its wrong too, but it always turns out to be correct

 

[Trebla]

Just to clarify,



Note that



holds if and only if x is non-negative (i.e. x > 0). If x is negative then the LHS of that statement is still well defined but does not equal the RHS. This is essentially where the definition of an absolute value comes in to account for this. (obviously we are assuming x is real)

Also, stop calling everyone 'retards' or 'noobs' especially when you just made a mistake yourself.

 

[laplace20]

Sorr, Mr I am doing 100% maths at uni and am no good for anything else

Finally a use for those pain in the ass absolute values inside the log when you integrate, to be honest in high school I just left it out lol

 

[Trebla]

lol, wrong again. Stop assuming without seeing the evidence. I don't do "100% maths" at uni. In fact I'm not taking any mathematics units of study right now. My degree is Comm/Sci(Adv). Mathematics is just one part within the science degree.

 

[cutemouse]

or better yet let C= ln(A)

You really only need to worry about that when you're doing applications to ODEs and you need to deal with initial conditions etc. Makes it a bit easier.