1)

∫ √(9-x^2)/x^2 dx

let x = 3sinu so that dx = 3cosu du

substitute x and dx reduces it to

∫ cot^2(u) du

= ∫ csc^2(u) – 1 du

= -cotu – u + C to put x back in

using x = 3sinu as a guide, draw a right triangle with

angle = u , opp = x and hyp = 3

by pythagoras this gives adj = √(9-x^2)

so cotu = √(9-x^2) /x

also u = arcsin(x/3)

so

∫ √(9-x^2)/x^2 dx

= -cotu – u + C

= -√(9-x^2) /x – arcsin(x/3) + C

2) this requires 2 substitutions

∫ x^3 /√(x^2-9) dx

let x = 3secu so that dx = 3secutanu du

substitute x and dx reduces it to

27 ∫ sec^4(u) du

= 27 ∫ (1 + tan^2(u) )sec^2(u) du

let v = tanu so that dv = sec^2(u) du ie du = dv /sec^2(u)

substitute v and du reduces it to

= 27 ∫ 1 + v^2 dv

= 27(v + (v^3) /3 ) + C then put u back in

= 27(tanu + (tan^3(u)) /3 ) + C then to put x back in

using x = 3secu as a guide, draw a right triangle with

angle = u , adj = 3 and hyp = x

by pythagoras this gives opp = √(x^2-9)

so tanu = √(x^2-9) /3

∫ x^3 /√(x^2-9) dx

= 27(tanu + (tan^3(u)) /3 ) + C

= 27( √(x^2-9) /3 + (√(x^2-9) /3)^3 /3) + C

= 9√(x^2-9) + (1/3)*(x^2-9)^3 + C

now drop the C and substitute your numbers for the answer

.