How do you integrate #int 1/((x+1)(x^2+2x+2))# using partial fractions?

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Answer 1 · 2016-11-22T06:58:12

The answer is #=ln(x+1)-1/2ln(x^2+2x+2)+C#

Let's start with the decomposition in partial fractions

#1/((x+1)(x^2+2x+2))=A/(x+1)+(Bx+C)/(x^2+2x+2)#

#=(A(x^2+2x+2)+(Bx+C)(x+1))/((x+1)(x^2+2x+2))#

So,
#1=A(x^2+2x+2)+(Bx+C)(x+1)#

Let #x=0#, #=>#, #1=2A+C#

Let #x=-1#, #=>#, #1=A#

coefficients of x^2, #0=A+B# #=>#, #B=-1#

#C=-1

#1/((x+1)(x^2+2x+2))=1/(x+1)+(-x-1)/(x^2+2x+2)#

#=1/(x+1)-(x+1)/(x^2+2x+2)#

#int(dx)/((x+1)(x^2+2x+2))=intdx/(x+1)-int(((x+1))dx)/(x^2+2x+2)#

#intdx/(x+1)=ln(x+1)#

For the second integral, we use a substitution

#u=x^2+2x+2#

#du=(2x+2)dx=2(x+1)dx#

#int(((x+1))dx)/(x^2+2x+2)=1/2int(du)/u=1/2lnu#

#=1/2ln(x^2+2x+2)#

Putting it all together,

#int(dx)/((x+1)(x^2+2x+2))=ln(x+1)-1/2ln(x^2+2x+2)+C#