Question

How do you find the indefinite integral of `int (x^2+2x+3)/(x^2+3x^2+9x)`?

Answer 1

`int(x^2+2x+3)/(x^2+3x^2+9x)dx=1/4x+1/16ln(4x+9)-2/3ln(|1/sqrt((8/9x+1)^2-1)+1/(8/9x+1)|)`

Explanation:

`I=int(x^2+2x+3)/(x^2+3x^2+9x)dx`
`=int(x^2+2x+3)/(4x^2+9x)dx=1/4int(4x^2+8x+12)/(4x^2+9x)dx`
`=1/4int((4x^2+9x)+x+12)/(4x^2+9x)dx`

`=1/4int1dx+1/4int(x+12)/(4x^2+9x)dx`

`=1/4x+1/4intx/(4x^2+9x)dx+1/4int12/(4x^2+9x)dx`

`=1/4x+1/16int4/(4x+9)dx+3/4int1/(x^2+9/4x)dx`

For the first integral, let `X=4x+9`
`dX=4dx`

So:

`I=1/4x+1/16int1/XdX+3/4int1/((x+9/8)^2-81/64)dx`

Now let `x+9/8=9/8sec(theta)`

`dx=9/8sec(theta)tan(theta)d theta`, and because `sec(theta)^2-1=tan(theta)^2`::

`I=1/4x+1/16ln(X)+3/4int(9/8sec(theta)tan(theta))/(81/64tan(theta)^2)d theta`

`=1/4x+1/16ln(X)+2/3intsec(theta)/tan(theta)d theta`

`=1/4x+1/16ln(X)+2/3int(1/cos(theta))/(sin(theta)/(cos(theta)))d theta`

`=1/4x+1/16ln(X)+2/3int1/sin(theta)d theta`

We have `int1/sin(theta)d theta=-ln(|cot(theta)+csc(theta)|)`, you can have the proof here.

`I=1/4x+1/16ln(X)-2/3ln(|cot(theta)+csc(theta)|)`

`=1/4x+1/16ln(4x+9)-2/3ln(|cot(theta)+csc(theta)|)`

Because `theta=sec^(-1)(8/9x+1)`, `cot(theta)=cos(theta)csc(theta)`, `cos(sec^(-1)(x))=1/x` and `csc(sec^(-1)(x))=1/sin(sec^(-1)(x))=x/sqrt(x^2-1)`
(here is a proof that `sin(sec^(-1)(x))=sqrt(x^2-1)/x`.),

`I=1/4x+1/16ln(4x+9)-2/3ln(|1/sqrt((8/9x+1)^2-1)+1/(8/9x+1)|)`

\0/ Here's our answer !