<p>help! thanks in advance.</p>
<p>what is the integral of
1/(x^2 + 1)</p>
<p>and </p>
<p>what is the derivative of</p>
<p>invtangent(4x)</p>
<p>invsin(3x)</p>
<p>(4x^2 - 5)^1/3</p>
<p>arctan x</p>
<p>4/(1+16x^2)</p>
<p>3/(sq rt of 1-9x^2)</p>
<p>16x/3 (4x^2 - 5) ^ (-2/3)</p>
<p>Is it right? I don't know. I didn't study.</p>
<p>crap! our teacher didnt teach us that........arcsin stuff</p>
<p>dont listen to me</p>
<p>For the first question, arctan x sounds right. That's pure memorization. As for the derivatives question, I'm kind of lost... heh... wow... I'm more screwed than a screw.</p>
<p>i dont get the ln stuff....</p>
<p>these are pretty much the only derivatives i dont know...hope it wont be on the no calc portion.</p>
<p>5ky, isn't it just arctan x?</p>
<p>lets just goto sleep and stop stressing</p>
<p>Yea, I'm pretty sure it's just arctan x. It wouldn't be 1/2 ln(x^2+1) because the derivative of the bottom (2x) isn't a multiple of the top (1). If it was just 2x+1 on the bottom, then the integral would be 1/2 ln(2x+1) + C.</p>
<p>haha oops, you guys are right. silly me.</p>
<p>thanks. also:</p>
<p>why is the integral of 1/8x = (ln8x)/8 + C ?</p>
<p>wouldnt it be 1/8x = 8x^-1 and integral of that is -8x^-2 ???</p>
<p>theripcurl, it's a u-sub</p>
<p>u = 8x
du = 8dx</p>
<p>so you have (1/8) times the integral of du/u
which = (1/8)(ln|u| + C)
which = (1/8)(ln|8x| + C)
which = (ln8x)/8 + C</p>
<p>(4x^2 - 5)^1/3</p>
<p>Isn't that (8x/3)(4x^2-5)^(-2/3)?</p>
<p>you mixed up integrals and derivatives, theripcurl. you add one to the power in the integral and subtract in derivatives. thus if you tried to integrate 8x^-1 like that you would get 8x^0.</p>
<p>chillaxin, yes.</p>