How do you find the integral of #int t/(t^4+16)#?

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Answer 1 · 2017-02-09T13:49:11

The answer is #=1/8arctan(t^2/4)+C#

We perform this integral by substitution

Let #u=t^2/4#

#du=2/4tdt=t/2dt#

#t^4+16=16u^2+16=16(u^2+1)#

Therefore,

#int(tdt)/(t^4+16)=int(2du)/(16(u^2+1))#

#=1/8int(du)/(u^2+1)#

Let #u=tantheta#

#du=sec^2theta d theta#

#1+tan^2theta=sec^2 theta#

So,

#1/8int(du)/(u^2+1)=1/8int(sec^2 theta d theta)/(sec ^2 theta)#

#=1/8intd theta=theta/8#

#=1/8arctan(u)#

Therefore,

#int(tdt)/(t^4+16)=1/8arctan(t^2/4)+C#