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Calculus1 8 Online
OpenStudy (anonymous):

Use the substitution formula to evaluate the integral. 3t/(5+t^2)^4 dt [-1,0]

OpenStudy (loser66):

\[\int_{-1}^{0}\frac{3t}{(5+t^2)^4}dt\]right?

OpenStudy (anonymous):

Yes!

OpenStudy (anonymous):

Can you help me with the steps @Loser66

OpenStudy (loser66):

let u = 5+t^2 --> du =?

OpenStudy (anonymous):

3t dt?

OpenStudy (loser66):

nope, du = 2t dt, but I need 3t dt, right? so?

OpenStudy (anonymous):

Oh! Derivative of U my bad, it would be 2t

OpenStudy (anonymous):

So? lol

OpenStudy (loser66):

ok, i give you one more step: du = 2t dt --> \(\frac {du}{2} = tdt\) right? but I need 3t dt , so, I time 3 both sides. I get \[\frac{3}{2}du = 3t dt\] OK??? got it? can step up?

OpenStudy (anonymous):

hmm?

OpenStudy (loser66):

what hmmm? don't understand? why ? quite clear, |dw:1376361663816:dw|

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