Integrate the following #int t/(t^4 +2) dt #?

Question

22767 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-19T16:35:21

#1/(2sqrt2)tan^-1(t^2/sqrt2)+C#

Let #t^2=sqrt2tantheta#. This implies that #2tdt=sqrt2sec^2thetad theta#. Then:

#I=1/2int(2tdt)/((t^2)^2+2)=1/2int(sqrt2sec^2thetad theta)/((sqrt2tantheta)^2+2)#

Continuing on and using #tan^2theta+1=sec^2theta#:

#I=1/sqrt2intsec^2theta/(2tan^2theta+2)d theta=1/(2sqrt2)intsec^2theta/sec^2thetad theta=1/(2sqrt2)intd theta#

We're working in terms of #theta#:

#I=1/(2sqrt2)theta+C#

From #t^2=sqrt2tantheta# we see that #theta=tan^-1(t^2/sqrt2)#:

#I=1/(2sqrt2)tan^-1(t^2/sqrt2)+C#

Answer 2 · 2017-02-19T23:20:13

I solved this way:
enter image source here