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

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