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.

  • $\begingroup$ Have you tried a trigonometric substitution? Have you tried to break the integrand up into 2 parts by partial fractions? $\endgroup$ Sep 6, 2019 at 19:33
  • 2
    $\begingroup$ Partial fraction decomposition then trivial integrals of $1/(ax+b)$ $\endgroup$ Sep 6, 2019 at 19:34
  • 1
    $\begingroup$ 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/… $\endgroup$
    – 6005
    Sep 6, 2019 at 19:42

3 Answers 3


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.

  • 1
    $\begingroup$ Thanks for the help! $\endgroup$
    – Masie
    Sep 6, 2019 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 $\frac18\int\csc tdt=\frac18\ln\left|\frac{\tan t}{1+\sec t}\right|+C=\frac{1}{16}\ln\left|\frac{2-x}{2+x}\right|+C$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.