the integral of the square-root of (x-1)2 = x2/2 - x + C
arctan(x)
pi squared
5x lolzz 8)
No, unless "a" happens to be equal to 0, or to 1.
sin squared
-1² = 1
1 over x2 - 4 is the multiplicative inverse of x2 minus four 1/x2 - 4
Multiply both sides by sin(1-cos) and you lose the denominators and get (sin squared) minus 1+cos times 1-cos. Then multiply out (i.e. expand) 1+cos times 1-cos, which will of course give the difference of two squares: 1 - (cos squared). (because the cross terms cancel out.) (This is diff of 2 squares because 1 is the square of 1.) And so you get (sin squared) - (1 - (cos squared)) = (sin squared) + (cos squared) - 1. Then from basic trig we know that (sin squared) + (cos squared) = 1, so this is 0.
The indefinite integral of (1/x^2)*dx is -1/x+C.
2x squared minus 5x minus 3 factored is (2x+1)(x-3).
(2x - 1)(4x + 1)