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Use the substitution formula to evaluate the integral. 3t/(5+t^2)^4 dt [-1,0]
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\[\int_{-1}^{0}\frac{3t}{(5+t^2)^4}dt\]right?
Yes!
Can you help me with the steps @Loser66
let u = 5+t^2 --> du =?
3t dt?
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nope, du = 2t dt, but I need 3t dt, right? so?
Oh! Derivative of U my bad, it would be 2t
So? lol
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?
hmm?
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what hmmm? don't understand? why ? quite clear, |dw:1376361663816:dw|
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