Partial fractions \int5/((x^2+2x+2)(x-1))?

I am getting results like this, are they ok so far?

A+C=0, B-A+2C=0, 2C-B=5

1 Answer
Narad T.
Apr 13, 2018

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

Explanation:

We need

int(u'(x)dx)/(u(x))=ln(u(x))+C

int(dx)/(x^2+1)=arctan(x)+C

Perform the decomposition into partial fractions

(5)/((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))

The denominatoras are the same, compare the numerators

5=A(x^2+2x+2)+(Bx+C)(x-1)

Let x=1, =>, 5=5A, =>, A=1

Let x=0, =>, 5=2A-C, =>, C=2A-5=2-5=-3

Coefficients of x^2,

0=A+B, =>, B=-A=-1

Therefore,

(5)/((x-1)(x^2+2x+2))=1/(x-1)+(-x-3)/(x^2+2x+2)

So,

int(5dx)/((x-1)(x^2+2x+2))=int(1dx)/(x-1)-int((x+3)dx)/(x^2+2x+2)

=ln(|x-1|)-I

And

I=int((x+3)dx)/(x^2+2x+2)=int((1/2(2x+2)+2)dx)/(x^2+2x+2)

=1/2int((2x+2)dx)/(x^2+2x+2)+2int(dx)/(x^2+2x+2)

=1/2ln(|x^2+2x+2|)+2I_1

I_1=int(dx)/(x^2+2x+2)=int(dx)/((x+1)^2+1)

=arctan(x+1)

Finally,

int(5dx)/((x-1)(x^2+2x+2))=ln(|x-1|)-1/2ln(|x^2+2x+2|)-2arctan(x+1)+C

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