How do you integrate $(x^2-1)/(x^2-16)$ using partial fractions?

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Answer 1 · 2016-12-02T06:41:31

The answer is $=x-15/8ln(∣x+4∣)+15/8ln(∣x-4∣)+C$

Firstly

$(x^2-1)/(x^2-16)=1+15/(x^2-16)$

$x^2-16=(x+4)(x-1)$

Now we can do the decomposition in partial fractions

$15/(x^2-16)=15/((x+4)(x-1))=A/(x+4)+B/(x-4)$

$=(A(x-4)+B(x+4))/((x+4)(x-1))$

Therefore,

$15=A(x-4)+B(x+4)$

Let $x=4$ , $=>$ , $15=8B$

$B=15/8$

Let $x=-4$ , $=>$ , $15=-8A$

$A=-15/8$

$(x^2-1)/(x^2-16)=1-15/(8(x+4))+15/(8(x-4)$

$int((x^2-1)dx)/(x^2-16)=int1dx-int(15dx)/(8(x+4))+int(15dx)/(8(x-4)$

$=x-15/8ln(∣x+4∣)+15/8ln(∣x-4∣)+C$

How do you integrate (x^2-1)/(x^2-16)(x^2-1)/(x^2-16) using partial fractions?