First, let's take a look at our denominator, #(x-1)(x^2+2x+5)# to ensure it is completely factored.
#(x-1)# is a linear factor; nothing more can be done.
#x^2+2x+5# is quadratic, let's see if we can factor it:
#b^2-4ac# where #b=2,c=5,a=1#, #=4-4(5)(1)<0#
Negative discriminant means this quadratic has no real roots, so it's factored as far as possible.
Now, recall that when decomposing a fraction, a linear factor in the denominator #(x+a)# shows up the following way:
#A/(x+a)# where #A# is a constant coefficient we must determine.
Recall that quadratic factors in the denominator show up the following way:
#(Ax+B)/(ax^2+bx+c)#
Where #A, B# are the constant coefficients we must determine.
Decomposing #(x^2+9x+30)/((x-1)(x^2+2x+5))#
yields
#(x^2+9x+30)/((x-1)(x^2+2x+5))=A/(x-1)+(Bx+C)/(x^2+2x+5)#
Let's add up the righthand side by using #(x-1)(x^2+2x+5)# as a common denominator:
#(x^2+9x+30)/((x-1)(x^2+2x+5))=(A(x^2+2x+5))/((x-1)(x^2+2x+5))+((Bx+C)(x-1))/((x-1)(x^2+2x+5))#
#(x^2+9x+30)/((x-1)(x^2+2x+5))=(A(x^2+2x+5)+(Bx+C)(x-1))/((x-1)(x^2+2x+5))#
Now, we can set our numerators equal:
#x^2+9x+30=A(x^2+2x+5)+(Bx+C)(x-1)#
We can determine #A# by setting #x=1:#
#1+9+30=A(1+2+5)+(Bx+C)(1-1)#
#40=8A#
#A=5#
We cannot determine #B# and #C# by this method, however, so let's multiply out everything on the right:
#x^2+9x+30=Ax^2+2Ax+5A+Bx^2-Bx+Cx-C#
Group terms with the same degree of #x# (or lack of #x#) together and factor out the common factor:
#x^2+9x+30=Ax^2+Bx^2+2Ax-Bx+Cx+5A-C#
#x^2+9x+30=x^2(A+B)+x(2A-B+C)+5A-C#
We now have the following equations:
#A+B=1# (Because #1# is the coefficient of #x^2# on the left)
#5A-C=30# (Because #30# is the constant term on the left)
We know #A=5#, so
#A+B=1->5+B=1->B=-4#
#5A-C=30->5(5)-C=30->25-C=30->C=-5#
Plugging #A=5,B=-4,C=-5# into #A/(x-1)+(Bx+C)/(x^2+2x+5)#
Yields
#5/(x-1)+(-4x-5)/(x^2+2x+5)#
Thus,
#int(x^2+9x+30)/((x+1)(x^2+2x+5)dx)=int(5/(x-1)+(-4x-5)/(x^2+2x+5))dx#
Split it up and factor out constants:
#5intdx/(x-1)+int(-4x-5)/(x^2+2x+5)dx#
Factor out #-1# from #-4x-5#
#5intdx/(x-1)-int(4x+5)/(x^2+2x+5)dx#
