Can you multiply a natural log term by a constant to take power inside log term? For Calculus 2 homework I must prove that $\displaystyle\int \frac{du}{u^2-a^2} = \frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|+C$
The instructor wants us to use trigonometric substitution to solve. 
My question is, may I take the term (acquired along the way):
\begin{align}
\ln\left|\frac{a+u}{\sqrt{u^2-a^2}}\right|
&=\frac{2}{2}\ln\left|\frac{a+u}{\sqrt{u^2-a^2}}\right|\\[3px]
&=\frac12\ln\left|\frac{a+u}{\sqrt{u^2-a^2}}\right|^2\\[3px]
&=\frac12\ln\left|\frac{(a+u)^2}{(\sqrt{u^2-a^2})^2}\right|
\end{align}
I realize the rules of logarithms state that $\log_ax^p = p\log_ax$
However, I can't tell if this is a valid operation or I merely have flawed logic in this case. Thank you!
 A: Your third line is OK.  You are indeed following the stated identity for logarithms, which is always correct -- maybe there's some corner case where it doesn't hold but it seems pretty robust for finite arguments.
However instead I would suggest you expand the integrand in partial fractions.
A: The substitution you can use in this case is
$$
u=\frac{a}{\cos2t}
$$
so
$$
\frac{1}{u^2-a^2}=\frac{\cos^22t}{a^2\sin^22t},
\qquad
du=\frac{2a\sin2t}{\cos^22t}
$$
and the integral becomes
$$
\frac{2}{a}\int\frac{1}{\sin2t}\,dt=
\frac{1}{a}\int\frac{\cos^2t+\sin^2t}{\sin t\cos t}\,dt=
\frac{1}{a}\int
  \left(\frac{\cos t}{\sin t}+\frac{\sin t}{\cos t}\right)\,dt=
\frac{1}{a}\ln\left|\frac{\sin t}{\cos t}\right|+C
$$
Now we can divide by $2$ and square inside the logarithm:
$$
\frac{1}{2a}\ln\left|\frac{\sin^2t}{\cos^2t}\right|=
\frac{1}{2a}\ln\left|\frac{1-\cos2t}{1+\cos2t}\right|=
\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|
$$
In your computation you have
$$
\ln\left|\frac{a+u}{\sqrt{u^2-a^2}}\right|=
\ln\sqrt{\frac{(u+a)^2}{u^2-a^2}}=
\ln\sqrt{\frac{u+a}{u-a}}=\frac{1}{2}\ln\frac{u+a}{u-a}
$$
but there seems to be something wrong in how you arrived to this. You're missing an absolute value under the square root, to begin with.
