How do you integrate #int (x+3) /( (x-2)x)# using partial fractions?

Question

Evelyn Agard · Accepted Answer

# 5/2ln|x - 2 | -3/2ln|x| + c#

The numerators will be constants because the denominator's factors are linear. hence # (x+3)/((x-2)x) = A/(x-2) + B/x # proceed to multiply by (x-2)x. x + 3 = Ax + B(x-2).....................(1) in order to obtain It is now necessary to determine the values of A and B. Keep in mind that the term containing A will be zero if x = 0, and the term containing B will be zero if x = 2. let x = 0 in (1) : 3 = -2B # rArr B = -3/2 # let x = 2 in (1) : 5 = 2A # rArr A = 5/2 # # rArr (x+3)/((x-2)x) = (5/2)/(x-2) -( 3/2)/x# #rArr int(x+3)/((x-2)x) dx = 5/2 intdx/(x-2) - 3/2intdx/x # # = 5/2ln|x-2| -3/2ln|x| + c# where the integration constant is denoted by c.

Christopher Agresta · Answer

undefined To integrate $\int \frac{x+3}{(x-2)x} \, dx$ using partial fractions, first express the integrand as $\frac{A}{x-2} + \frac{B}{x}$. Then, find the values of $A$ and $B$ by equating numerators and solving for $A$ and $B$. Once you have the partial fraction decomposition, you can integrate each term separately and combine the results to get the final integral.