How do you find the antiderivative of #int 1/(x^2+3x) dx#?

1 Answer
Dec 26, 2016

#1/3ln|x|-1/3ln|x+3|+C#.

Explanation:

Write down:
#=int1/(x(x+3))dx#
#=int( { }/x + { }/(x+3))dx#.

Then apply the cover-up rule for partial fractions. To find out what goes over the #x#, use your finger to cover up the factor #x# in the denominator of the fraction on the first line, and replace all other #x#'s with zero. Similarly, to find out what goes over the #x+3#, cover up the #x+3# and replace the other #x# with #-3#. In each case, you replace #x# with whatever value of #x# makes the expression under your finger zero.

#=int( (1)/(0+3))/x + (1/(-3))/(x+3)dx#
#=1/3int1/x-1/(x+3)dx#.