Write down:
=int1/(x(x+3))dx=∫1x(x+3)dx
=int( { }/x + { }/(x+3))dx=∫(x+x+3)dx.
Then apply the cover-up rule for partial fractions. To find out what goes over the xx, use your finger to cover up the factor xx in the denominator of the fraction on the first line, and replace all other xx's with zero. Similarly, to find out what goes over the x+3x+3, cover up the x+3x+3 and replace the other xx with -3−3. In each case, you replace xx with whatever value of xx makes the expression under your finger zero.
=int( (1)/(0+3))/x + (1/(-3))/(x+3)dx=∫10+3x+1−3x+3dx
=1/3int1/x-1/(x+3)dx=13∫1x−1x+3dx.
