Question

`int(x^2+x+2)/{(x-1)(x-2)}dx` with respect to x ?

Answer 1

`x+4{-ln|(x-1)|+2ln|(x-2)|}+C, or`

`x+4ln((x-2)^2/|(x-1)|)+C`.

Explanation:

Let, `I=int(x^2+x+2)/((x-1)(x-2))dx=int(x^2+x+2)/(x^2-3x+2)dx`.

Observe that, the degree of the polynomials in Numerator

and Denominator is the same, so the rational integrand is

Improper.

So, first, we have to make it Proper. To this end, the long

division is, usually, performed. Here, we proceed as under :

`x^2+x+2=(x^2-3x+2)+4x.`

`:. I=int{(x^2-3x+2)+4x}/(x^2-3x+2)dx`,

`=int{(x^2-3x+2)/(x^2-3x+2)+(4x)/(x^2-3x+2)}dx`,

`=int1dx+4intx/((x-1)(x-2))dx`,

`=x+4I_1`, where,

`I_1=intx/((x-1)(x-2))dx`.

Here, we will decompose `x/((x-1)(x-2))` using

Method of Partial Fraction.

We let, `x/((x-1)(x-2))=A/(x-1)+B/(x-2); A,B in RR`.

Then, using Heavyside's Method to determine `A,B`, we have,

`A=[x/(x-2)]_(x=1)=1/(1-2)=-1`.

`B=[x/(x-1)]_(x=2)=2/(2-1)=2`.

`:. I_1=int{-1/(x-1)+2/(x-2)}dx`,

`=-ln|(x-1)|+2ln|(x-2)|`.

`rArr I=x+4{-ln|(x-1)|+2ln|(x-2)|}+C, or`

`I=x+4ln((x-2)^2/|(x-1)|)+C`.

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