Are these integrations the same? For the integration of: 
$$v^2 + 6k^2 = -Mv \frac{dv}{dx}$$
I rearranged to get:
$$\int \frac{1}{-M} dx = \frac{1}{2} \int \frac{2v}{v^2 + 6k^2} dv$$
Is this the same as the following integral that is in the book?
$$\int \frac{1}{M} dx = -\frac{1}{2} \int \frac{2v}{v^2 + 6k^2} dv$$
I wonder because my final answer ends up being:
$$\frac{x}{-M} = \frac{1}{2} \ln(\frac{v^2 + 6k^2}{10k^2})$$
But the book gives:
$$\frac{x}{M} = \frac{1}{2} \ln(\frac{10k^2}{v^2 + 6k^2})$$
Have I done something wrong or is it the same thing?
 A: hint take minus inside inside the ln on other side it will make the  reciprocal of your answer giving the final answer of your solution and note in integration we can always shift the constants.
A: They are the same answer. If you take the minus over to the right, and incorporate it into the $\ln$ function, it becomes $1/x$.
A: On the left hand side, the negative can be removed from the denominator as a constant negative one. That constant can be pulled out of the integral, then moved to the right hand side. Using the property of logarithms that allows a constant factor to become the power of the inside, then the stuff inside the logarithm becomes inverted. So the answers are equivalent. 
A: $$v(x)^2+6k^2=-mv(x)v'(x)\Longleftrightarrow$$
$$v'(x)=-\frac{6k^2+v(x)^2}{mv(x)}\Longleftrightarrow$$
$$-\frac{mv(x)v'(x)}{6k^2+v(x)^2}=1\Longleftrightarrow$$
$$\int-\frac{mv(x)v'(x)}{6k^2+v(x)^2}\space\text{d}x=\int1\space\text{d}x\Longleftrightarrow$$
$$\int-\frac{mv(x)v'(x)}{6k^2+v(x)^2}\space\text{d}x=x+\text{C}$$

Now, for the intergal, substitute $u=6k^2+v(x)^2$ and $\text{d}u=2v(x)v'(x)\space\text{d}x$:
$$\int-\frac{mv(x)v'(x)}{6k^2+v(x)^2}\space\text{d}x=-\frac{m}{2}\int\frac{1}{u}\space\text{d}u=-\frac{m\ln\left|u\right|}{2}+\text{C}=-\frac{m\ln\left|6k^2+v(x)^2\right|}{2}+\text{C}$$

So, we get that:
$$-\frac{m\ln\left|6k^2+v(x)^2\right|}{2}=x+\text{C}\Longleftrightarrow\left|6k^2+v(x)^2\right|=\text{C}\exp\left[-\frac{2x}{m}\right]$$
