Calculus
posted by Kyle .
How do you integrate:
1/(x^21) dx
?

draw a right triangle, x as the hypotensue, 1 as the adjacent side to angle theta, and the opposite side is sqrt (x^21)
costheta=1/x
sintheta* dT=1/x^2 dx or
dx= x^2 sinTheta dT
the integral is now...
INt 1/tanTheta*x^2 sinTheta dT
INT cosT/SinT* sec^2T*sinT dT
INT= secT dT=ln(secT+tanT)
= ln(x+x^21)
check all that, I did it in my head. 
Hm, I've never done integrals using that method before. I'll check it.