Integration using logs

Jun 2009
186
5
Mars
Hello I am having trouble with this question.

INtegrate:
f(2x+5)/(x+4) dx

My attempt:
1/2x+5f 1/x+4 dx
2x+5 Ln (x+4) + c
 

galactus

MHF Hall of Honor
Jul 2006
3,002
1,124
Chaneysville, PA
\(\displaystyle \int\frac{2x+5}{x+4}dx=2x-3ln(x+4)+C\)

You're close. But there should be a minus where the plus is, and a 3 instead of a 5 in front of the ln.
 
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Jun 2009
186
5
Mars
But how do you get the -3. That is the part I cannot do.
By the way thanks for the help and nice avatar thing.
 

galactus

MHF Hall of Honor
Jul 2006
3,002
1,124
Chaneysville, PA
If we expand, we get

\(\displaystyle \frac{2x+5}{x+4}=2-\frac{3}{x+4}\)

Now, integrate that and you can see why.
 
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Dec 2009
3,120
1,342
Alternatively,

\(\displaystyle \int{\frac{2x+5}{x+4}}dx=\int{\frac{x+4+x+1}{x+4}}dx=\int{\left(1+\frac{x+1}{x+4}\right)}dx\)

\(\displaystyle u=x+4\)

\(\displaystyle x+1=u-3\)

\(\displaystyle \int{\left(1+\frac{u-3}{u}\right)}du=\int{\left(1+1-3u^{-1}\right)}du=\int{\left(2-3u^{-1}\right)}du\)

\(\displaystyle =2u-3ln|u|+C=2(x+4)-3ln|x+4|+C\)
 
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