AB calculus problem

<p>how do you derive ln(x+4+e^-3x)</p>

<p>d/dx[lnx] = (1/x)<em>(x')
d/dx[ln(x+4+e^-3x) = 1/(x+4+e^-3x)</em>(1+(-3e^-3x))
d/dx[ln(x+4+e^-3x) = (1-3e^-3x)/(x+4+e^-3x)</p>

<p><a href="1%20-%203e%5E(-3x)">1/(x+4+e^(-3x))</a></p>

<p>:rolleyes:</p>

<p>Haha, I just did this problem yesterday.</p>

<p>how do you take the antiderivite of sin(x^2)</p>

<p>tables or integrals.wolfram.com</p>

<p>This is easy~</p>

<p>Use MacClaurin Polynomials.</p>

<p>Sin X = x - (x^3 / 3!) + (x^5 / 5!) - ...
Sin X^2 = x^2 - (x^6 / 3!) + (x^10 / 5!) - ... + (-1)^(n+1)(x^(4n-2) / (2n - 1)!)</p>

<p>Answer: x^3 / 3 - (x^7 / 7x3!) + (x^11 / 11x5!) - ... + (-1)^(n+1)(x^(4n-1)) / (4n-1)(2n-1)!) = antiderivative of sin(x^2)</p>

<p>But then again... this is AB, so crap, use a calculator. lol</p>