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

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.

• Have you tried a trigonometric substitution? Have you tried to break the integrand up into 2 parts by partial fractions? – Bernard Massé Sep 6 '19 at 19:33
• Partial fraction decomposition then trivial integrals of $1/(ax+b)$ – Peter Foreman Sep 6 '19 at 19:34
• Partial fractions is the best way to solve it. See if you can find a partial fractions example in your textbook (if you have one). Otherwise, here's the solution to a few example problems that might help: math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/… – 6005 Sep 6 '19 at 19:42

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.

• Thanks for the help! – Masie Sep 6 '19 at 20:00

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 $$+C$$.

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