Hello,
I am having trouble integrating the following equation, I thought I could do it but it is not the same as the answer in the back of the book. Can someone help me?
The integral of (2x+3)/(x+1)dx thanks
$\displaystyle \int\frac{2x+3}{x+1}\,dx$
$\displaystyle =\int\frac{2x+2+1}{x+1}\,dx$
$\displaystyle =\int\frac{2(x+1)+1}{x+1}\,dx$
$\displaystyle =\int\left[\frac{2(x+1)}{x+1}+\frac1{x+1}\right]dx$
$\displaystyle =\int\left(2+\frac1{x+1}\right)dx$
You could have also used polynomial long division.
$\displaystyle \int{\frac{2x+3}{x+1}} dx$
$\displaystyle = 2 \int{ \frac{x}{x+1} }dx + 3 \int{ \frac{1}{x+1} }dx$
$\displaystyle = 2 \int{ \frac{x+1-1}{x+1} }dx + 3 \ln{(x+1)} + C$
$\displaystyle
= 2 \left( \int{ \frac{x+1}{x+1} }dx + \int{ \frac{-1}{x+1} }dx \right) + 3 \ln{(x+1)} + C$
$\displaystyle = 2 \left( \int{ 1 }dx + \int{ \frac{-1}{x+1} }dx \right) + 3 \ln{(x+1)} + C$
$\displaystyle = 2 \left( x - \ln{(x+1)} \right) + 3 \ln{(x+1)} + C$
$\displaystyle = 2x + \ln{(x+1)} + C$