1. ## Basic integral problem.

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

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

2. Originally Posted by navynukejones
Is the answer to integrate x/x+3dx this:

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

3. ## Are you sure?

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.

4. 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. $\displaystyle \int{\frac{dx}{x+3}}=\ln|x+3|+C$

5. 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 $\displaystyle \int \frac{x}{x+3} \, dx = \int \frac{x}{u} \, du$

But hey you have to get rid of that x in the above expression
Thats why write $\displaystyle \frac{x}{x+3} = 1 - \frac{3}{x+3}$. Both the terms now are easily integrable

6. Originally Posted by navynukejones
Is the answer to integrate x/x+3dx this:

ln|x+3|+c ????
Ok..just to let you know...$\displaystyle \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$$\displaystyle =x-c\ln|x+c|+C$