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

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

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Answer 1 · 2018-08-02T23:32:30

$\int \frac{x^{2} + 2 x + 3}{x^{2} + 3 x^{2} + 9 x} d x = \frac{1}{4} x + \frac{1}{16} ln (4 x + 9) - \frac{2}{3} ln ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ∣ ∣ ∣ ∣ ∣ ∣ \frac{1}{\sqrt{{(\frac{8}{9} x + 1)}^{2} - 1}} + \frac{1}{\frac{8}{9} x + 1} ∣ ∣ ∣ ∣ ∣ ∣ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$ $int(x^2+2x+3)/(x^2+3x^2+9x)dx=1/4x+1/16ln(4x+9)-2/3ln(|1/sqrt((8/9x+1)^2-1)+1/(8/9x+1)|)$

Explanation:

$I = \int \frac{x^{2} + 2 x + 3}{x^{2} + 3 x^{2} + 9 x} d x$ $I=int(x^2+2x+3)/(x^2+3x^2+9x)dx$
$= \int \frac{x^{2} + 2 x + 3}{4 x^{2} + 9 x} d x = \frac{1}{4} \int \frac{4 x^{2} + 8 x + 12}{4 x^{2} + 9 x} d x$ $=int(x^2+2x+3)/(4x^2+9x)dx=1/4int(4x^2+8x+12)/(4x^2+9x)dx$
$= \frac{1}{4} \int \frac{(4 x^{2} + 9 x) + x + 12}{4 x^{2} + 9 x} d x$ $=1/4int((4x^2+9x)+x+12)/(4x^2+9x)dx$

$= \frac{1}{4} \int 1 d x + \frac{1}{4} \int \frac{x + 12}{4 x^{2} + 9 x} d x$ $=1/4int1dx+1/4int(x+12)/(4x^2+9x)dx$

$= \frac{1}{4} x + \frac{1}{4} \int \frac{x}{4 x^{2} + 9 x} d x + \frac{1}{4} \int \frac{12}{4 x^{2} + 9 x} d x$ $=1/4x+1/4intx/(4x^2+9x)dx+1/4int12/(4x^2+9x)dx$

$= \frac{1}{4} x + \frac{1}{16} \int \frac{4}{4 x + 9} d x + \frac{3}{4} \int \frac{1}{x^{2} + \frac{9}{4} x} d x$ $=1/4x+1/16int4/(4x+9)dx+3/4int1/(x^2+9/4x)dx$

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

So:

$I = \frac{1}{4} x + \frac{1}{16} \int \frac{1}{X} d X + \frac{3}{4} \int \frac{1}{{(x + \frac{9}{8})}^{2} - \frac{81}{64}} d x$ $I=1/4x+1/16int1/XdX+3/4int1/((x+9/8)^2-81/64)dx$

Now let $x + \frac{9}{8} = \frac{9}{8} sec (θ)$ $x+9/8=9/8sec(theta)$

$d x = \frac{9}{8} sec (θ) tan (θ) d θ$ $dx=9/8sec(theta)tan(theta)d theta$ , and because ${sec (θ)}^{2} - 1 = {tan (θ)}^{2}$ $sec(theta)^2-1=tan(theta)^2$ ::

$I = \frac{1}{4} x + \frac{1}{16} ln (X) + \frac{3}{4} \int \frac{\frac{9}{8} sec (θ) tan (θ)}{\frac{81}{64} {tan (θ)}^{2}} d θ$ $I=1/4x+1/16ln(X)+3/4int(9/8sec(theta)tan(theta))/(81/64tan(theta)^2)d theta$

$= \frac{1}{4} x + \frac{1}{16} ln (X) + \frac{2}{3} \int \frac{sec (θ)}{tan (θ)} d θ$ $=1/4x+1/16ln(X)+2/3intsec(theta)/tan(theta)d theta$

$= \frac{1}{4} x + \frac{1}{16} ln (X) + \frac{2}{3} \int \frac{\frac{1}{cos (θ)}}{\frac{sin (θ)}{cos (θ)}} d θ$ $=1/4x+1/16ln(X)+2/3int(1/cos(theta))/(sin(theta)/(cos(theta)))d theta$

$= \frac{1}{4} x + \frac{1}{16} ln (X) + \frac{2}{3} \int \frac{1}{sin (θ)} d θ$ $=1/4x+1/16ln(X)+2/3int1/sin(theta)d theta$

We have $\int \frac{1}{sin (θ)} d θ = - ln (| cot (θ) + csc (θ) |)$ $int1/sin(theta)d theta=-ln(|cot(theta)+csc(theta)|)$ , you can have the proof here.

$I = \frac{1}{4} x + \frac{1}{16} ln (X) - \frac{2}{3} ln (| cot (θ) + csc (θ) |)$ $I=1/4x+1/16ln(X)-2/3ln(|cot(theta)+csc(theta)|)$

$= \frac{1}{4} x + \frac{1}{16} ln (4 x + 9) - \frac{2}{3} ln (| cot (θ) + csc (θ) |)$ $=1/4x+1/16ln(4x+9)-2/3ln(|cot(theta)+csc(theta)|)$

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

$I = \frac{1}{4} x + \frac{1}{16} ln (4 x + 9) - \frac{2}{3} ln ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ∣ ∣ ∣ ∣ ∣ ∣ \frac{1}{\sqrt{{(\frac{8}{9} x + 1)}^{2} - 1}} + \frac{1}{\frac{8}{9} x + 1} ∣ ∣ ∣ ∣ ∣ ∣ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$ $I=1/4x+1/16ln(4x+9)-2/3ln(|1/sqrt((8/9x+1)^2-1)+1/(8/9x+1)|)$

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

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