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,

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

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