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.
