integral from 0 to 1 of 8x/((4x^2 + 1)^2) using substitution? I understand that u should = 4x^2 + 1
Then change the limits for u to 1 to 5 by filling the original limits in for u. And du = 8x dx. So I'm left with u^-2. What now?

Question

integral from 0 to 1 of 8x/((4x^2 + 1)^2) using substitution? I understand that u should = 4x^2 + 1
Then change the limits for u to 1 to 5 by filling the original limits in for u. And du = 8x dx. So I'm left with u^-2. What now?

anonymous · Answer

find the anti derivative using the power rule backwards

anonymous · Answer

Okay antiderivative is 1/2 x^-1 then

anonymous · Answer

$$\int u^ndu=\frac{u^{n+1}}{n+1}$$ in your case $n=-2$

anonymous · Answer

not quite

anonymous · Answer

good question. Atleast you did not ask "AY YO I HAVE A QUESTION"

anonymous · Answer

u= 4x^2+1

anonymous · Answer

should be $$\int u^{-2}du=-u^{-1}=-\frac{1}{u}$$

anonymous · Answer

evaluate at 5, 1, and subtract

jhannybean · Answer

Are you concurrently taking Calc 3 and Calc 4 together?

anonymous · Answer

wait no that's not right is it. Okay so it's actually - 1/u. Okay that makes sense. Then I fill in the new limits and subtract?

anonymous · Answer

calc 4? is that before calc 5, 6, and 7?

anonymous · Answer

yes

anonymous · Answer

Alright that works. Thanks satellite73