how do i integrate the problem:
\int (x^2+2x+1)/(2x+3) dx
I'm so stuck on this
Thank You
Hello, larryboi7!
$\displaystyle \int\frac{x^2+2x+1}{2x+3}\,dx$
Note that the numerator is a quadratic, the denominator is linear.
. . The fraction is "improper".
Use long division: .$\displaystyle (x^2+2x+1) \div (2x+3) \;=\;\frac{1}{2}\,x + \frac{1}{4} + \frac{\frac{1}{4}}{2x+3}$
The integral becomes: .$\displaystyle \int\left(\frac{1}{2}\,x + \frac{1}{4} + \frac{1}{4}\!\cdot\!\frac{1}{2x+3}\right)dx \;=\;\frac{1}{2}\!\int\! x\,dx + \frac{1}{4}\!\int\! dx + \frac{1}{4}\!\int\!\frac{dx}{2x+3}$
And you can finish it, right?