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

Answer 1

Emilia Aguilar

#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#

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Answer 2

Scarlett Allison

I solved this way:

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Answer 3

Charles Arocha

To integrate the function ( \int \frac{t}{t^4 + 2} , dt ), you can use the substitution method. Let ( u = t^2 ). Then ( du = 2t , dt ). Making the substitution, the integral becomes ( \frac{1}{2} \int \frac{1}{u^2 + 2} , du ). This integral can be solved by using the inverse tangent function, so the final result is ( \frac{1}{4} \tan^{-1} \left( \frac{t^2}{\sqrt{2}} \right) + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

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