How would you go about integrating $\frac{1}{(-16+4x^2)}$ with respect to $x$?

Question

How would you go about integrating $\frac{1}{(-16+4x^2)}$ with respect to $x$?

I've tried U-substition and integration by parts, but I can't see to get an answer and I'm assuming I'm missing some sort of trick:

When I used U substition, i've tried letting $u=4x^2-16$, but the $du$ term always has an "x" in it.

When I used integration by parts, i pulled out a $1/4$ and my two parts were $\frac{1}{x-2}$ and $\frac{1}{x+2}$, but I couldn't see a way of simplifying from there.

I guess I can try partial fractions, but again, I dont see an way of making it simpler, and i'm not so comfortable with that method.

Any help is appreciated.

Answer 1

Hint: If you factor the denominator, you should get something like $$ \frac{1}{a(x - r)(x + s)} $$ Once you do this, you can apply Partial-Fraction Decomposition to finish the solution. Are you familiar with it?

Edit: Here are some partial fractions example integrals, just like the one you posted here.

Answer 2

With $x=2\sec t$, it's $\frac18\int\csc tdt=\frac18\ln\left|\frac{\tan t}{1+\sec t}\right|+C=\frac{1}{16}\ln\left|\frac{2-x}{2+x}\right|+C$.

Answer 3

Start off with \begin{align} \int\frac1{-16+4x^2}\ dx&=\frac14\int\frac1{x^2-4}\ dx\\ \end{align}

We want to decompose

\begin{align} \frac1{x^2-4}&=\frac{A}{x-2}+\frac{B}{x+2}\\ &=\frac{A(x+2)+B(x-2)}{x^2-4}\\ &=\frac{Ax+2A+Bx-2B}{x^2-4}\\ &=\frac{(A+B)x+(2A-2B)}{x^2-4} \end{align}

The numerator must match up, so we have the set of equations $$\left\{ \begin{align*} A+B=0 \\ 2A-2B=1\end{align*} \right.$$ The solution to the above is $(A,B)=(\frac14,-\frac14)$.

As such: \begin{align} \int\frac1{-16+4x^2}\ dx&=\frac14\int\frac1{x^2-4}\ dx\\ &=\frac14\int\frac{\frac14}{x-2}+\frac{-\frac14}{x+2}\ dx\\ &=\frac1{16}\int\frac1{x-2}-\frac1{x+2}\ dx\\ &=\frac1{16}\bigg(\ln|x-2|-\ln|x+2|\bigg)+C\\ &=\frac1{16}\ln\bigg|\frac{x-2}{x+2}\bigg|+C \end{align}

Don't forget the $+C$.