Original post by Ano123
Integral of arctan(3x)/[x(1+x^2)] between x=0 and x=infinity.
...
Edit: Have you written down the question correctly? Wolfram Alpha doesn't give a nice answer
.Edit: I input it wrong. It's solvable.
(edited 8 years ago)
Original post by B_9710
πln2 
How?
Original post by Kurwaaaa
How?
How what?
Original post by B_9710
How what?
DId you get the answer, I presume
.
Original post by Kurwaaaa
How?
kurwa how
Original post by Kvothe the arcane
DId you get the answer, I presume .
.Differentiation under the integral sign.
Original post by B_9710
Differentiation under the integral sign.
Ah, one of those techniques. Thank you
. Any chance you can help the OP?Original post by TeeEm
easy??
Compared to hard integral thread?
Original post by B_9710
Compared to hard integral thread?
definitely not A level
Hmmm...I have (x*arctan(3x)-1/6(ln(1+9x^2)))*ln(x)*arctan(x)
Where did I go wrong? And how do I do the limits?
Where did I go wrong? And how do I do the limits?
(edited 8 years ago)
Original post by Username002
Hmmm...I have (x*arctan(3x)-1/6(ln(1+9x^2)))*ln(x)*arctan(x)
Where did I go wrong? And how do I do the limits?
Where did I go wrong? And how do I do the limits?
What integration technique have you used?
I split the integral up. Can I even do that? I split it up into three integrals.
arctan(3x), 1/x and 1/(1+x^2)
arctan(3x), 1/x and 1/(1+x^2)
(edited 8 years ago)
Original post by Username002
I split the integral up. Can I even do that? I split it up into three integrals.
arctan(3x), 1/x and 1/(1+x^2)
arctan(3x), 1/x and 1/(1+x^2)
You cannot do that, the integrand is a product of those 3 terms so you cannot split it up in this way.
If I may ask what level of maths are you currently working at?
(edited 8 years ago)
A level, why?
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