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

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- Feb 24th 2009, 08:21 AMjarnyintegration
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 - Feb 24th 2009, 08:37 AMReckoner
$\displaystyle \int\frac{2x+3}{x+1}\,dx$

$\displaystyle =\int\frac{2x+2+1}{x+1}\,dx$

$\displaystyle =\int\frac{2(x+1)+1}{x+1}\,dx$

$\displaystyle =\int\left[\frac{2(x+1)}{x+1}+\frac1{x+1}\right]dx$

$\displaystyle =\int\left(2+\frac1{x+1}\right)dx$

You could have also used polynomial long division. - Feb 24th 2009, 08:38 AMjanvdl
$\displaystyle \int{\frac{2x+3}{x+1}} dx$

$\displaystyle = 2 \int{ \frac{x}{x+1} }dx + 3 \int{ \frac{1}{x+1} }dx$

$\displaystyle = 2 \int{ \frac{x+1-1}{x+1} }dx + 3 \ln{(x+1)} + C$

$\displaystyle

= 2 \left( \int{ \frac{x+1}{x+1} }dx + \int{ \frac{-1}{x+1} }dx \right) + 3 \ln{(x+1)} + C$

$\displaystyle = 2 \left( \int{ 1 }dx + \int{ \frac{-1}{x+1} }dx \right) + 3 \ln{(x+1)} + C$

$\displaystyle = 2 \left( x - \ln{(x+1)} \right) + 3 \ln{(x+1)} + C$

$\displaystyle = 2x + \ln{(x+1)} + C$ - Feb 24th 2009, 10:19 AMKrizalid
Of course janvdl meant to say $\displaystyle \ln|x+1|.$ :)

- Feb 24th 2009, 10:33 AMjarny
thanks a lot guys

- Feb 25th 2009, 01:49 AMjanvdl