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,

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

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