How do you integrate #int (3(4y²-7y-12))/(y(y+2)(y-3))# using partial fractions?

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Answer 1 · 2017-01-13T07:37:07

The answer is #=6ln(∣y∣)+27/5ln(∣y+2∣)+3/5ln(∣y-3∣)+C#

Perform the decomposition into partial fractions

#(3(4y^2-7y-12))/(y(y+2)(y-3))=A/(y)+B/(y+2)+C/(y-3)#

#=(A(y+2)(y-3)+B(y)(y-3)+C(y)(y+2))/(y(y+2)(y-3))#

Therefore, equalising the numerators

#12y^2-21y-36=A(y+2)(y-3)+B(y)(y-3)+C(y)(y+2)#

Let #y=0#, #=>#, #-36=-6A#, #=>#, #A=6#

Let #y=-2#,#=>#, #48+42-36=10B#, #=>#, #B=54/10=27/5#

Let #y=3#, #=>#, #108-63-36=15C#, #=>#, #C=9/15=3/5#

So,

#(3(4y^2-7y-12))/(y(y+2)(y-3))=6/(y)+(27/5)/(y+2)+(3/5)/(y-3)#

#int(3(4y^2-7y-12)dy)/(y(y+2)(y-3))=6intdy/(y)+27/5intdy/(y+2)+3/5intdy/(y-3)#

#=6ln(∣y∣)+27/5ln(∣y+2∣)+3/5ln(∣y-3∣)+C#