For my math class I need to solve a series of equations (all of which deal with antidifferentiation), now I understand the simple stuff such as: ∫2x dx=2x^2/2=>∫2x=x^2, and I also understand how to solve derivatives with the form ∫(ax+b)^r dx=((ax+b)^(r+1))/(a*(r+1)), however, when I have to solve derivatives with the special case rule, I get a little confused: rule ∫(ax+b)^-1 dx=1/a*ln(|ax+b|)+c, where ln is a natural logarithm, ln=loge..

Well anyway, I can solve them when the equation is straight forward, however,my teacher did not go over dealing with derivatives of the form ((ax+b)^2)/x... How do integrate this form of a derivative? the same also applies for the form x/(ax+b)^2, can someone please help me with this? Also, if you know any math terms in esperanto could you also chuck them in? Like I think 1+1=2 is something like unu plus unu estas du or unu plus unu egalas du..

Anyways thanks in advance! Dankon por via helpo!

Saluton!
(1)
∫(((ax+b)^2)/x)dx = ∫((ax)^2 + 2abx + b^2)/x dx
= ∫a^2*x dx + ∫2ab dx + ∫b^2/x dx
= (ax)^2/2 + 2abx + b^2*ln(x) + konstanto //

(2)
∫x/(ax+b)^2 dx .
{Do you know how to work with the substitution? Basicly you just choose which part you can change for u (in this case it`s ax+b)and then derivate it to get du (in this case the derivate of ax+b is a)}
Anstataŭo: [u = ax+b => du = adx, dx = du/a, x= (u-b)/a]
=> ∫x/(ax+b)^2 dx = ∫((u-b)/a)/u^2 *du/a
= ∫(u-b)/(au)^2 du = ∫1/(a^2*u) du - ∫b/(au)^2 du
= ln(u)/a^2 + (b*1/u)/a^2 + K1
anstataŭo: u = ax+b
= ( ln(ax+b) + b/*(ax+b) )/a^2 + konstanto //

Vi povas viziti http://www.wolframalpha.com/ por klarigi viajn demandojn.

Dankon por via helpo!!

Substitution comes in very handy, so learn it well.