Integration using logs

**Awsom Guy** · May 21st 2010, 01:12 AM

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** · May 21st 2010, 02:53 AM

$\int\frac{2x+5}{x+4}dx=2x-3ln(x+4)+C$

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.

**Awsom Guy** · May 21st 2010, 03:08 AM

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** · May 21st 2010, 03:19 AM

If we expand, we get

$\frac{2x+5}{x+4}=2-\frac{3}{x+4}$

Now, integrate that and you can see why.

**Archie Meade** · May 21st 2010, 03:48 AM

Alternatively,

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

$u=x+4$

$x+1=u-3$

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

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

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