How do you integrate #\int _ { 3} ^ { 4} \frac { 3x ^ { 2} + 10x } { x ^ { 2} - 4} d x#?

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Answer 1 · 2017-07-22T14:14:45

The answer is #=3x+2ln(|x+2|)+8ln(|x-2|)+C#

First, we simplify the function

#(3x^2+10x)/(x^2-4)=3+(10x+12)/(x^2-4)#

#(10x+12)/(x^2-4)=(2(5x+6))/((x+2)(x-2))#

Let's perform the decompositio into partial fractions

#(5x+6)/((x+2)(x-2))=A/(x+2)+B/(x-2)=(A(x-2)+B(x+2))/((x+2)(x-2))#

The denominators are the same, we compare the numerators

#5x+6=A(x-2)+B(x+2)#

Let #x=-2#, #=>#, #-4=-4A#, #=>#, #A=1#

Let #x=2#, #=>#, #16=4B#, #B=4#

Therefore,

#(3x^2+10x)/(x^2-4)=3+2/(x+2)+8/(x-2)#

#int((3x^2+10x)dx)/(x^2-4)=int3dx+int(2dx)/(x+2)+int(8dx)/(x-2)#

#=3x+2ln(|x+2|)+8ln(|x-2|)+C#