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

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Answer 1 · 2018-08-02T23:32:30

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

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

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

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

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

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 !

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