Solution
The correct answer is 12alog|x−ax+a| + C
Explanation
∫dxx2−a2
∫dx(x−a)(x+a)
∫A(x−a).dx+∫B(x+a).dx ---- (i)
Solving for values of A and B, we get
1=A(x+a)+B(x−a)
Putting x=a, we get A=12a
Putting x=−a, we get A=−12a
Substituting these values in (i),
∫12a(x−a).dx−∫12a(x+a).dx
12a[∫1(x−a).dx−∫1(x+a).dx]
12a[log|x−a|−log|x+a|] + C
12alog|x−ax+a| + C