Basic integral problem.

**navynukejones** · April 22nd 2008, 08:54 AM

Is the answer to integrate x/x+3dx this:

ln|x+3|+c ????

**Isomorphism** · April 22nd 2008, 09:01 AM

Originally Posted by navynukejones

Is the answer to integrate x/x+3dx this:

ln|x+3|+c ????

No it is,
$\int \frac{x}{x+3} \, dx = \int (1 - \frac{3}{x+3}) \, dx = x - 3\ln|x+3| + C$

**navynukejones** · April 22nd 2008, 09:40 AM

I think you have to use the formula: the integral of du/u = ln|u|+C and in this problem u=x+3 and therefore du=dx.

**disclaimer** · April 22nd 2008, 09:44 AM

Originally Posted by navynukejones

I think you have to use the formula: the integral of du/u = ln|u|+C and in this problem u=x+3 and therefore du=dx.

It works only when the numerator is the derivative of the denominator, eg. $\int{\frac{dx}{x+3}}=\ln|x+3|+C$

**Isomorphism** · April 22nd 2008, 09:51 AM

Originally Posted by navynukejones

I think you have to use the formula: the integral of du/u = ln|u|+C and in this problem u=x+3 and therefore du=dx.

Just observe carefully.... there is a x term in the numerator. So according to you u = x+3, du = dx.

So $\int \frac{x}{x+3} \, dx = \int \frac{x}{u} \, du$

But hey you have to get rid of that x in the above expression

Thats why write $\frac{x}{x+3} = 1 - \frac{3}{x+3}$ . Both the terms now are easily integrable

**Mathstud28** · April 22nd 2008, 02:12 PM

Originally Posted by navynukejones

Is the answer to integrate x/x+3dx this:

ln|x+3|+c ????

Ok..just to let you know... $\int\frac{x}{x+c}dx=\int\frac{x+c-c}{x+c}dx=\int\frac{x+c}{x+c}dx-\int\frac{c}{x+c}dx$ $=x-c\ln|x+c|+C$

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