Here is what I have so far:

$\displaystyle

\int ln(x^2+10x+16)~dx = xln(x^2+10x+16) - \int \frac {2x^2+10}{x^2+10x+16}~dx

$

I am stuck after this though...

- Sep 7th 2008, 07:54 PMredman223[SOLVED] Integration by parts with a polynomial fraction inside?
Here is what I have so far:

$\displaystyle

\int ln(x^2+10x+16)~dx = xln(x^2+10x+16) - \int \frac {2x^2+10}{x^2+10x+16}~dx

$

I am stuck after this though... - Sep 7th 2008, 08:04 PM11rdc11
long division of the polynomial

- Sep 7th 2008, 08:12 PMredman223
How would this divide out? We have not really covered long division inside an integral in class yet. I know it splits into two integrals though right?

- Sep 7th 2008, 08:21 PM11rdc11
$\displaystyle 2 + \bigg(\frac{- 20x - 22}{x^2 +10x +16}\bigg)$

I would like to show you how to do the long divison but not sure how to display it on latex. - Sep 7th 2008, 08:46 PMShyam
$\displaystyle

=x \ln(x^2+10x+16) - \int \frac {2x^2+20+32-10x-22}{x^2+10x+16}~dx

$

$\displaystyle

=x \ln(x^2+10x+16) - \int \left(2-\frac {10x+22}{x^2+10x+16}\right)~dx

$

$\displaystyle =x \ln(x^2+10x+16) -2x + \int \frac {5(2x+10)-28} {x^2+10x+16} ~dx $

$\displaystyle =x \ln(x^2+10x+16) -2x +5 \ln(x^2+10x+16) -28 \int \frac {1} {(x+5)^2-3^2} ~dx $

$\displaystyle =x \ln(x^2+10x+16) -2x +5 \ln(x^2+10x+16) - \frac{28}{2\times 3} \ln \left( \frac {x+5+3} {x+5-3} \right)$

$\displaystyle =x \ln(x^2+10x+16) -2x +5 \ln(x^2+10x+16) - \frac{14}{3} \ln \left( \frac {x+8} {x+2} \right)$ - Sep 8th 2008, 12:26 AMchoovuck
there is a mistake in the first line written by redman: the numerator of the last integral should be $\displaystyle x(2x+10)$, not $\displaystyle 2x^2+10$

then $\displaystyle \frac{2x^2+10x}{x^2+10x+16}=\frac{2(x^2+10x+16)-10x-32}{x^2+10x+16}$ and so on... - Sep 8th 2008, 12:39 AM11rdc11
You could also use partial fractions or trig substitution, whatever you feel more comfortable with.

- Jan 10th 2010, 03:15 PMJhevon
actually, i'd approach it this way

$\displaystyle \int \ln (x^2 + 10x + 16)~dx = \int \ln [(x + 2)(x + 8)]~dx = \int \ln (x + 2) ~dx + \int \ln (x + 8)~dx$

Now in light of the fact that we know $\displaystyle \int \ln x ~dx = x \ln x - x + C$ (we can get this via integration by parts, but you really should know this by heart), each of those two integrals are really easy with a simple linear substitution

Haha, I just realized the date on this thing. really late, but it might help someone anyway....