[SOLVED] Integrating log function

• Feb 24th 2010, 04:06 PM
Archduke01
[SOLVED] Integrating log function
http://euler.vaniercollege.qc.ca/web...8ca5108931.png

my $\displaystyle u = (x^2+1)$, $\displaystyle du = 2x$

$\displaystyle 4 \int (2x+1)/(x^2+1)$

I would like to turn that into du/u but there's a + 1 in the numerator, which isn't part of du's value. What am I supposed to do?
• Feb 24th 2010, 04:11 PM
ione
Quote:

Originally Posted by Archduke01

my $\displaystyle u = (x^2+1)$, $\displaystyle du = 2x$

$\displaystyle 4 \int (2x+1)/(x^2+1)$

I would like to turn that into du/u but there's a + 1 in the numerator, which isn't part of du's value. What am I supposed to do?

Try breaking up the fraction into 2 fractions

$\displaystyle \frac{8x+4}{x^2+1}=\frac{8x}{x^2+4}+\frac{4}{x^2+4 }$
• Feb 24th 2010, 04:22 PM
Archduke01
Thank you!
• Feb 24th 2010, 04:30 PM
$\displaystyle \int{\frac{2x+1}{x^2+1}}dx=\int{\frac{2x}{x^2+1}dx +\int{\frac{1}{x^2+1}}}dx$
$\displaystyle \int{\frac{2x}{x^2+1}}dx=\int{\frac{du}{u}}$