Integrate: { -3/x +6/x^2 + 3/(x+3) dx
(With { dx being the notation for integration, I can't find the actual sign, and with 2 over 1 as the x substitutes)

Question

Integrate: { -3/x +6/x^2 + 3/(x+3) dx
(With { dx being the notation for integration, I can't find the actual sign, and with 2 over 1 as the x substitutes)

anonymous · Answer

So is that
$$\int_1^2\left(-\frac{3}{x}+\frac{6}{x^2}+\frac{3}{x+3}\right)\,dx~~?$$

katielong · Answer

Yes

anonymous · Answer

For the first and third terms, you would use the logarithm, since
$$\int \frac{dx}{x}=\ln|x|+C$$
The second integral uses the power rule: For $n\not=-1$,
$$\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$$

katielong · Answer

Okay so it should be:
[-3lnx - 6x^-1 + 3ln(x+3) ]   ?

anonymous · Answer

That's right. Now you need to evaluate the endpoints as you would with any definite integral.

If $F(x)=-3\ln|x|-\dfrac{6}{x}+3\ln|x+3|+C$, find $F(2)-F(1)$.

katielong · Answer

Okay, thank you. 
= (-3ln2-3+3ln5)-(-3ln1-6+3ln4)
How would this be simplified? It supposedly goes to 3-2ln(8/5) ?

anonymous · Answer

Close, that 2 should be a 3.