#intx^2/(a^2-x^2)^(3/2)dx#
rewrite as follows
#=int(x^2-a+a)/(a^2-x^2)^(3/2)dx#
#=color(red)(int(x^2-a^2)/(a^2-x^2)^(3/2)dx)+color(blue)(inta^2/(a^2-x^2)^(3/2)dx)---(1)#
we will deal with each integral in turn, and leave the constant until the end
from #(1)" " #the red integral
#color(red)(int(x^2-1)/(a^2-x^2)^(3/2)dx)#
#color(red)(I_1=-int(a^2-x^2)/(a^2-x^2)^(3/2)dx=-int1/(a^2-x^2)^(1/2)dx#
we proceed by substitution
#color(red)(x=asinu=>dx=acosudu)#
#:. color(red)(I_1=-int1/(cancel((a^2(1-sin^2x))^(1/2)))xx cancel(acosu)du#
#color(red)(I_1=-intdu#
#color(red)(I_1=-u=-sin^(-1)(x/a)---(2)#
now take the second (blue )integral from #(2)" "~ and solve by substitution
#color(blue)(I_2=inta^2/(a^2-x^2)^(3/2)dx#
#color(blue)(x=asinu=>dx=acosudu)#
#color(blue)(I_2=inta^2/((a^2(1-sin^2u))^(3/2)) xx a cosudu#
which simplifies to
#color(blue)(intdu/cos^2u=intsec^2udu=tanu#
#"now "sinu=x/a#
#:.tanu=(x/a)/(sqrt(a^2-x^2)/a)=x/(sqrt(a^2-x^2)#
#color(blue)(I_2=x/(sqrt(a^2-x^2))#
the final integral becomes
#x/(sqrt(a^2-x^2))-sin^(-1)(x/a)+c#