ive gotten the easy ones, but these ones are throwing me for a loop...
we got these...
integrate
(x-3)dx/(x^3+3x^2+2x)
x^2 dx(x^2+2x+2)^2
(sin^2 x) dx / (1 + sin^2 x)
Well for the first step, notice that
$\displaystyle \frac{A}{A+B}=1-\frac{B}{A+B}$.
For the second one I just used $\displaystyle \cos^2 x + \sin^2 x = 1$.
I can tell you that the integral is quite difficult. The answer involves both arctangent and hyperbolic arctangent!
the first one
$\displaystyle \frac{x-3}{x^3+3x^2+2x}$ = $\displaystyle \frac{A}{x}+\frac{B}{x+2}+\frac{C}{x+1}$
$\displaystyle \frac{x-3}{x^3+3x^2+2x}$ =$\displaystyle \frac{(Ax^2+3Ax+2A)+(Bx^2+Bx)+(Cx^2+Cx)}{x^3+3x^2+ 2x}$$\displaystyle =\frac{(A+B+C)x^2 + (3A+B+C)x + 2A}{x^3+3x^2+2x}$
solve
$\displaystyle (A+B+C)=0$
$\displaystyle (3A+B+C)=1$
$\displaystyle 2A=-3$
Two polynimials are equal if and only the coefficients of like powers are equal
The coefficients of the cubed terms are the same
the coefficients of the squared terms are the same
etc
In your example there is no cubic term on the left So the coefficient of the cubic term on the right hand side namely A must be 0
The coefficient of the squared term on the left is 1 so the coefficient
of the squared term on the right namely 2A + B =1
since A= 0 B must be one
and so on