solving $\int \frac{2x+3}{9x^2-12x+8}$

Question

$$\int \frac{2x+3}{9x^2-12x+8}dx$$

first $9x^2-12x+8=-(3x-2)^2+4$

$$\int \frac{2x+3}{4+(3x-2)^2}dx=\frac{1}{4} \int \frac{2x+3}{1+(\frac{3x-2}{2})^2}dx$$

$u=\frac{3x-2}{2}$

$du=\frac{3}{2}dx$

$$\frac{1}{6}\int \frac{2x+3}{1+(u)^2}du=\frac{1}{6}\int \frac{2x}{1+(u)^2}du+\frac{1}{6}\int \frac{3}{1+(u)^2}du=\frac{1}{3}\int \frac{x}{1+(u)^2}du+\frac{1}{2}\int \frac{1}{1+(u)^2}du$$

$s=1+u^2$

$ds=2udu$

$$\frac{1}{6}\int \frac{1}{s}ds+\frac{1}{2}\int \frac{1}{1+(u)^2}du$$

$$=\frac{1}{6}\ln|1+\Big(\frac{3x-2}{2}\Big)^2|+\frac{1}{2}\arctan\Big(\frac{3x+2}{2}\Big)+c$$

I cant find where it got wrong

Answer 1

Mathematica gives:

$\frac{1}{9} \log \left(9 x^2-12 x+8\right)+\frac{13}{18} \tan ^{-1}\left(\frac{1}{2} (3 x-2)\right)$

Answer 2

$$\frac{1}{6}\int \frac{2x+3}{1+(u)^2}du=\frac{1}{6}\int \frac{2\color{green}x}{1+(u)^2}du+\frac{1}{6}\int \frac{3}{1+(u)^2}du=\frac{1}{\color{red}6}\int \frac{2\color{green}x}{1+(u)^2}du+\frac{1}{2}\int \frac{1}{1+(u)^2}du$$

But in subsequent working you need to get rid of the $\color{green}x$ term in the integral

Answer 3

$\int\frac{2x+3}{9x^2-12+8}dx=\int\frac{2x-\frac{4}{3}}{9x^2-12+8}dx+\int\frac{\frac{13}{3}}{9x^2-12+8}dx=\frac{1}{9}\int\frac{18x-12}{9x^2-12+8}dx+\frac{13}{3}\int\frac{1}{9x^2-12+8}dx=\frac{1}{9}\int\frac{18x-12}{9x^2-12+8}dx+\frac{13}{3}\int\frac{1}{(3x-2)^2+4}dx$

now setting $$u=9x^2-12x+8 \to du=(18x-12)dx$$ $$s=3x-2\to ds=3dx$$ you get
$\frac{1}{9}\int\frac{du}{u}+\frac{13}{9}\int\frac{ds}{s^2+4}=\frac{\log(u)}{9}+\frac{13}{36}\int\frac{ds}{\frac{s^2}{4}+1}$ $$t=\frac{s}{2}\to dt=\frac{ds}{2}$$ $\frac{\log(u)}{9}+\frac{13}{18}\int\frac{ds}{p^2+1}=\frac{\log(u)}{9}+\frac{13}{18}\int\frac{dt}{t^2+1}=\frac{\log(u)}{9}+\frac{13}{18}\arctan(t)=$
$$\frac{\log(9x^2-12x+8)}{9}+\frac{13}{18}\arctan(\frac{3x}{2}-1)+C$$