Question

How do you integrate `int(x+1)/((2x-4)(x+5)(x-2))` using partial fractions?

Answer 1

First we observe that

`int(x+1)/((2x-4)(x+5)(x-2))dx=1/2*int (x+1)/[(x+5)*(x-2)^2]dx`

Hence we have to find constants A,B,C such as

`(x+1)/[(x+5)*(x-2)^2]=A/(x+5)+B/(x-2)+C/(x-2)^2`

We can calculate these constants by giving x three different values

for example `x=0,x=-1,x=3`

hence we get

`(x+1)/[(x+5)*(x-2)^2]=-2/[49*(x+5)]+2/[49*(x-2)]+3/[14(x-2)^2]`

Hence now we have that

#int(x+1)/((2x-4)(x+5)(x-2))dx=1/2*int (x+1)/[(x+5)*(x-2)^2]dx= 1/2*[int (-2)/(49*(x+5))dx +int 2/(49(x-2))dx+int 3/[(14)(x-2)^2]dx]= 1/2*[-2/49*ln(x+5)+2/49*ln(x-2)-3/14*(1/(x-2))]+c#