Lucky for us, the denominator in the integral is nice and factored for us, so our partial fraction decomposition will be of the form #A/(x+3)+B/(x-2)+C/(x-7)#.
Start by breaking #(2x+1)/((x+3)(x-2)(x-7))# into partial fractions:
#(2x+1)/((x+3)(x-2)(x-7))=A/(x+3)+B/(x-2)+C/(x-7)#
Multiplying through by #(x+3)(x-2)(x-7)# gives:
#2x+1=A(x-2)(x-7)+B(x+3)(x-7)+C(x+3)(x-2)#
That equation is looking pretty nasty, but note that if we set #x=2#, we get:
#2(2)+1=A((2)-2)((2)-7)+B((2)+3)((2)-7)+C((2)+3)((2)-2)#
#5=A(0)(-5)+B(5)(-5)+C(5)(0)#
#5=-25B#
#B=-1/5#
Likewise, set #x=7# to get:
#2(7)+1=A((7)-2)((7)-7)+B((7)+3)((7)-7)+C((7)+3)((7)-2)#
#15=A(5)(0)+B(10)(0)+C(10)(5)#
#15=50C#
#C=15/50=3/10#
And finally, set #x=-3# to find #A#:
#2(-3)+1=A((-3)-2)((-3)-7)+B((-3)+3)((-3)-7)+C((-3)+3)((-3)-2)#
#-5=A(-5)(-10)+B(0)(-10)+C(0)(-5)#
#-5=50A#
#A=-5/50=-1/10#
So we have #A=-1/10#, #B=-1/5#, and #C=3/10#. Plugging these back into our original equation, we see:
#(2x+1)/((x+3)(x-2)(x-7))=(-1/10)/(x+3)+(-1/5)/(x-2)+(3/10)/(x-7)#
#color(white)(XX)=-1/(10(x+3))-1/(5(x-2))+3/(10(x-7))#
Now all that's left is to integrate this:
#int(2x+1)/((x+3)(x-2)(x-7))dx=int-1/(10(x+3))-1/(5(x-2))+3/(10(x-7))dx#
#color(white)(XX)=int-1/(10(x+3))dx-int1/(5(x-2))dx+int3/(10(x-7))dx#
#color(white)(XX)=-1/10int1/(x+3)dx-1/5int1/(x-2)dx+3/10int1/(x-7)dx#
#color(white)(XX)=-1/10lnabs(x+3)-1/5lnabs(x-2)+3/10lnabs(x-7)+C#
And we're done.
