How do you evaluate the integral #int dx/((2x-1)(x+2))#?
1 Answer
Use partial fractions. See below.
Explanation:
The denominator is already factored.
Multiply through by the denominator of the left-hand side:
#(2x-1)(x+2)[1/cancel((2x-1)(x+2))=A/(cancel(2x-1))+B/cancel(x+2)]#
We need to solve for
#1=B(-4-1)#
#1=-5B#
#B=-1/5#
#1=A(1/2+2)#
#1=A(5/2)#
#A=2/5#
We put these values back into our partial fractions and replace this as the integrand.
Technically you should use a substitution before integrating. Split up the integral.
For the first integral,
For the second integral,
Integrate.
Substitute back in.
Note: the absolute value signs account for the domain of the natural log function (
#x>0# ).
By rules of logarithms, this is equivalent to
#1/5ln(|(2x-1)/(x+2)|)+C#