# How to solve the indefinite integral of `(2x^2 +20x+44)/ (x^2 +7x+ 12)` ? `int(2x^2 +20x+44)/ (x^2 +7x+ 12)dx`

*print*Print*list*Cite

`int(2x^2 +20x+44)/ (x^2 +7x+ 12)dx`

Since both numerator and denominator have a degree of 2, we may simplify the integrand by dividing them.

`2`

`x^2+7x+12``bar(| 2x^2+20x+44)`

`-` `2x^2+14x+24`

`-----------`

`6x + 20`

So the integral becomes:

`int(2x^2 +20x+44)/ (x^2 +7x+ 12)dx= int(2 + (6x+20)/(x^2+7x+12))dx`

Since the integrand is sum of two functions, then it can be express as sum of two integrals.

`= int 2dx + int (6x+20)/(x^2+7x+12) dx`

For the first integral, apply the formula `int kdx=kx+C` .

`= 2x+C+ int(6x+20)/(x^2+7x+12)dx`

For the second integral, since the degree of numerator is less than that of denominator, expand the integrand using partial fractions.

So, factor the denominator and express it as two fractions.

`(6x+20)/((x+3)(x+4))=A/(x+3)+B/(x+4)`

Then, determine the values of A and B. To do so, multiply both sides by the LCD.

`(6x+20)/((x+3)(x+4))*(x+3)(x+4)=(A/(x+3)+B/(x+4))*(x+3)(x+4)`

`6x+20=A(x+4)+B(x+3)`

To solve for A, set x=-3,

`6(-3)+20=A(-3+4)+B(-3+3)`

`-18+20=A(1)+B(0)`

`2=A`

And to solve for B, set x=-4.

`6(-4)+20=A(-4+4)+B(-4+3)`

`-24+20=A(0)+B(-1)`

`-4=-B`

`4=B`

Hence, the second integral becomes:

`2x+C+ int(6x+20)/(x^2+7x+12)dx`

`=2x+C+int(2/(x+3)+4/(x+4))dx`

Again, express the integral as sum of two integrals.

`=2x+C+int2/(x+3)dx + int4/(x+4)dx`

`= 2x+ C +2int 1/(x+3)dx+4int1/(x+4)dx`

Then, apply the formula `int 1/u du = ln u + C` .

`=2x+C+2ln(x+3)+C+4ln(x+4)+C`

Since C represents any number, we can represent the sum of the three C's as C only.

`=2x+2ln(x+3)+ 4ln(x+4)+C`

**Hence, **

`int(2x^2 +20x+44)/ (x^2 +7x+ 12)=2x+2ln(x+3)+ 4ln(x+4)+C` .