Question

Integrate t/(t^4 + 3). Help! Should I try breaking this up .. or substitution..?

Answer 1

Let \(u=t^2\), then \(\dfrac{1}{2}du=dt\):
\[\frac{1}{2}\int\frac{dt}{u^2+3}\]
Then a trig sub should work.

Answer 2

should be \(du\) in the integral...

Answer 3

Try the following:
\[\int\limits_{}{}\frac{ t }{ t^4+3 }dt=\frac{ \sqrt{3} }{ 6 } \int\limits_{}{}\frac{\frac{ 2}{ \sqrt{3} } t}{ \left( \frac{ t^2 }{ \sqrt{3}} \right)^2+1 }dt=\frac{ \sqrt{3} }{ 6 }\int\limits_{}{}\frac{ d \left( \frac{ t^2 }{ \sqrt{3} } \right) }{ \left( \frac{ t^2 }{ \sqrt{3} } \right)^2+1}\]Does it look familiar to you?

Answer 4

Something like\[\frac{ \sqrt{3} }{ 6 }\arctan \left( \frac{ t^2 }{ \sqrt{3}} \right)+C\]perhaps?

Answer 5

As you see, rearranging things you can have an immediate integral

Answer 6

Ohhhh, thanks both for your help.