# How to find integrals like int (1-x)(x^2-4)dx

amcavoy
I already know the answer to this, but would like your opinion on the quickest way to calculate integrals like these. I am finding myself evaluating similar integrals in my differential equation homework, and the method that I used (parts and partial fractions) is a mess and takes way too long (if it were to appear on a test).

$$\int\left(\frac{1-x}{x^2-4}\right)dx$$

Thanks.

## Answers and Replies

Gold Member
In general, partial fractions is the only way to go. In this particular case, you can split the integral up like this to do it pretty fast:
$$\int-\frac{x-1}{x^2-4}dx$$
$$=-\int\frac{x-2}{(x+2)(x-2)}+\frac{2}{2(x+2)(x-2)}dx$$
$$=-\int\frac{1}{x+2}+\frac{2+x}{2(x+2)(x-2)}-\frac{x}{2(x^2-4)}}dx$$
$$=-\int\frac{1}{x+2}+\frac{1}{2(x-2)}-\frac{x}{2(x^2-4)}}dx$$
Now you can directly compute the first two parts as logs and the last as a log by recognizing x as one quarter the derivative of the denomenator

amcavoy
LeonhardEuler said:
In general, partial fractions is the only way to go. In this particular case, you can split the integral up like this to do it pretty fast:
$$\int-\frac{x-1}{x^2-4}dx$$
$$=-\int\frac{x-2}{(x+2)(x-2)}+\frac{2}{2(x+2)(x-2)}dx$$
$$=-\int\frac{1}{x+2}+\frac{2+x}{2(x+2)(x-2)}-\frac{x}{2(x^2-4)}}dx$$
$$=-\int\frac{1}{x+2}+\frac{1}{2(x-2)}-\frac{x}{2(x^2-4)}}dx$$
Now you can directly compute the first two parts as logs and the last as a log by recognizing x as one quarter the derivative of the denomenator

Well it looks like the only way to save time here is to use my TI-89 .

Thanks for the reply Euler.