1. ## Integration question

2. Note $\frac{4x^2 + 3x + 1}{x(2x + 1)} = \frac{1}{x} - \frac{1}{2x + 1} + 2$.

3. How do you know to break it up like this?

4. 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.

5. Originally Posted by fterh
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 $u = 2x+1$.

6. Thanks guys, I've figured out a way to do it already.