# Integration question

• Oct 21st 2010, 02:34 AM
fterh
Integration question
• Oct 21st 2010, 02:40 AM
NOX Andrew
Note $\frac{4x^2 + 3x + 1}{x(2x + 1)} = \frac{1}{x} - \frac{1}{2x + 1} + 2$.
• Oct 21st 2010, 02:52 AM
fterh
How do you know to break it up like this?
• Oct 21st 2010, 03:05 AM
NOX Andrew
For integrands involving improper polynomial fractions, it's a common technique to rewrite the integrand as a proper polynomial fraction. Improper polynomial fractions are fractions in which the highest degree in the numerator is greater than or equal to the highest degree in the denominator.
• Oct 21st 2010, 06:36 AM
TheCoffeeMachine
Quote:
Judging by the hint, I think here is the way they want you to do it:

Split the integral into two:

$\displaystyle \int_{1}^{2}\frac{4x^2+3x+1}{x(2x+1)}\;{dx} = \displaystyle \underbrace{\int_{1}^{2}\frac{4x^2+3x}{x(2x+1)}\;{ dx}}_{I_{1}}+ \underbrace{\int_{1}^{2}\frac{1}{x(2x+1)}\;{dx}}_{ I_{2}}$

Since we are given that $\displaystyle \left[\ln\left(\frac{x}{x(2x+1)}\right)\right]' = \frac{1}{x(2x+1)}$, we have $\displaystyle I_{2} = \bigg[\ln\left(\frac{x}{x(2x+1)}\right)\bigg]_{1}^{2}$.

For $\displaystyle I_{1}$, we write $\displaystyle I_{1} = \int_{1}^{2}\frac{4x^2+3x}{x(2x+1)}\;{dx} = \int_{1}^{2}\frac{x(4x+3)}{x(2x+1)}\;{dx} = \int_{1}^{2}\frac{4x+3}{2x+1}\;{dx}$

Then let $u = 2x+1$.
• Oct 21st 2010, 06:37 AM
fterh
Thanks guys, I've figured out a way to do it already.