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

Answer 1

Luke Alvarez

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#

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Answer 2

Charles Arocha

To find the integral of ( \int \frac{t}{t^4 + 16} , dt), you can use the substitution method. Let ( u = t^2 ), then ( du = 2t , dt ). Rewrite the integral with respect to ( u ) as follows: [ \frac{1}{2} \int \frac{1}{u^2 + 16} , du ] This integral can be evaluated using the inverse tangent function. The integral becomes: [ \frac{1}{2} \cdot \frac{1}{4} \arctan \left( \frac{u}{4} \right) + C ] Replace ( u ) with ( t^2 ) and simplify to get the final answer.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.