# Intergrate Calc dont you love it

• Mar 3rd 2008, 05:34 PM
Bust2000
Intergrate Calc dont you love it
Intergrate (2x^2+4x+22) / (x^2+2x+10) dx
• Mar 3rd 2008, 05:58 PM
Soroban
Hello, Bust20001

Quote:

$\displaystyle \int\frac{2x^2+4x+22}{x^2+2x+10}\,dx$

Divide: . $\displaystyle \int\left(2 + \frac{2}{x^2 +2x+10}\right)\,dx$

. . . . . $\displaystyle =\;\;\int\left(2 + \frac{2}{(x+1)^2+9}\right)\,dx$

. . . . . $\displaystyle = \;\;2x + \frac{2}{3}\arctan\left(\frac{x+1}{3}\right)+ C$

• Mar 3rd 2008, 05:58 PM
mr fantastic
Quote:

Originally Posted by Bust2000
Intergrate (2x^2+4x+22) / (x^2+2x+10) dx

$\displaystyle \frac{2x^2 + 4x + 22}{x^2 + 2x + 10} = 2\left( \frac{x^2 + 2x + 11}{x^2 + 2x + 10} \right) = 2\left( \frac{(x^2 + 2x + 10) + 1}{x^2 + 2x + 10} \right)$

$\displaystyle = 2\left( 1 + \frac{1}{x^2 + 2x + 10} \right) = 2\left( 1 + \frac{1}{(x + 1)^2 + 9} \right) = \left( 1 + \frac{1}{3} \, \frac{3}{(x + 1)^2 + 3^2} \right)$.

You should have better luck with this form .....
• Mar 3rd 2008, 06:09 PM
roy_zhang
Quote:

Originally Posted by Bust2000
Intergrate (2x^2+4x+22) / (x^2+2x+10) dx

When your integrand is in the form $\displaystyle \frac{P(x)}{Q(x)}$, where $\displaystyle P(x),Q(x)$ are both polynomials, and the order of $\displaystyle P(x)$ is greater or equal to the order of $\displaystyle Q(x)$; always try long division first before integrating.

Here we have:
$\displaystyle \frac{2x^2+4x+22}{x^2+2x+10} = 2+\frac{2}{x^2+2x+10}$

Now,
$\displaystyle \int\frac{2x^2+4x+22}{x^2+2x+10}dx = \int (2+\frac{2}{x^2+2x+10})dx = \int2dx + \int[\frac{2}{(x+1)^2+3^2}]dx$

use $\displaystyle u$ substitution (let $\displaystyle u= x+1$)

we got:

$\displaystyle \int \frac{2x^2+4x+22}{x^2+2x+10}dx=2x + \frac{2}{3}\arctan(\frac{x+1}{3}) + C$