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.

 

[LeonhardEuler]

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]

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.