Firstly
(x^2-1)/(x^2-16)=1+15/(x^2-16)x2−1x2−16=1+15x2−16
x^2-16=(x+4)(x-1)x2−16=(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)15x2−16=15(x+4)(x−1)=Ax+4+Bx−4
=(A(x-4)+B(x+4))/((x+4)(x-1))=A(x−4)+B(x+4)(x+4)(x−1)
Therefore,
15=A(x-4)+B(x+4)15=A(x−4)+B(x+4)
Let x=4x=4, =>⇒, 15=8B15=8B
B=15/8B=158
Let x=-4x=−4, =>⇒, 15=-8A15=−8A
A=-15/8A=−158
(x^2-1)/(x^2-16)=1-15/(8(x+4))+15/(8(x-4)x2−1x2−16=1−158(x+4)+158(x−4)
int((x^2-1)dx)/(x^2-16)=int1dx-int(15dx)/(8(x+4))+int(15dx)/(8(x-4)∫(x2−1)dxx2−16=∫1dx−∫15dx8(x+4)+∫15dx8(x−4)
=x-15/8ln(∣x+4∣)+15/8ln(∣x-4∣)+C=x−158ln(∣x+4∣)+158ln(∣x−4∣)+C
