Integrate x^2+2x+3/x^3+3x^2+9x

Question

anonymous · Answer

try a u - sub

anonymous · Answer

$$y=x^3+3x^2+9x\
du=(3x^2+6x+9)dx\
\frac{du}{3}=(x^2+2x+3)dx$$ pretty well cooked this one

misty1212 · Answer

HI!!

misty1212 · Answer

you got this after the u substitution?  you are going to end up with 
$$\frac{1}{3}\int \frac{du}{u}$$ which you solve in one step
`

solomonzelman · Answer

sate, just that you wrote y=... and then du=... very good approach though

anonymous · Answer

Thank you!

solomonzelman · Answer

I mean,

$\large\color{slate}{  u=x^3+3x^2+9x                   }$
$\large\color{slate}{  du=(3x^2+6x+9)dx                  }$
$\large\color{slate}{  (1/3)du=(x^2+x+3)dx                  }$

solomonzelman · Answer

(just a little variable err)

solomonzelman · Answer

camila, and then make sure to switch back to x, as you are done

anonymous · Answer

If you're feeling particularly ambitious you can also split up the given expression into partial fractions:
$$\frac{x^2+2x+3}{x^3+3x^2+9x}=\frac{x^2+2x+3}{x(x^2+3x+9)}=\frac{A}{x}+\frac{Bx+C}{x^2+3x+9}$$