How do I use partial fractions to evaluate the integral?

#int_1^2(4y^2-6y-12)/(y(y+2)(y-3)#dy

I keep on getting 4ln(2) - 8/5ln(3), and it's not right.

1 Answer
George C.
Sep 19, 2017

#int_1^2 (4y^2-6y-12)/(y(y+2)(y-3)) dy = 8/5(3 ln 2 - ln 3) ~~ 1.5693#

Explanation:

#(4y^2-6y-12)/(y(y+2)(y-3)) = A/y+B/(y+2)+C/(y-3)#

We can find #A#, #B# and #C# using Oliver Heaviside's cover up method:

#A = (4(color(blue)(0))^2-6(color(blue)(0))-12)/(((color(blue)(0))+2)((color(blue)(0))-3)) = (-12)/(-6) = 2#

#B = (4(color(blue)(-2))^2-6(color(blue)(-2))-12)/((color(blue)(-2))((color(blue)(-2))-3)) = 16/10 = 8/5#

#C = (4(color(blue)(3))^2-6(color(blue)(3))-12)/((color(blue)(3))((color(blue)(3))+2)) = 6/15 = 2/5#

So:

#int_1^2 (4y^2-6y-12)/(y(y+2)(y-3)) dy#

#=int_1^2 (2(1/y)+8/5(1/(y+2))+2/5(1/(y-3))) dy#

#=[ 2ln abs(y)+8/5 ln abs(y+2)+2/5 ln abs(y-3) ]_1^2#

#=(2ln abs(color(blue)(2))+8/5 ln abs(color(blue)(2)+2)+2/5 ln abs(color(blue)(2)-3) ) - (2ln abs(color(blue)(1))+8/5 ln abs(color(blue)(1)+2)+2/5 ln abs(color(blue)(1)-3))#

#=(2 ln 2+8/5 ln 4+0) - (0 + 8/5 ln 3 + 2/5 ln 2)#

#=(2 + 16/5 - 2/5)ln2 - 8/5 ln 3#

#=24/5 ln 2 - 8/5 ln 3#

#=8/5 ln 8 - 8/5 ln 3#

#=8/5 ln(8/3)#

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