p. 582

2.

∫(3/(Sqrt[1 + 9  (y^2)])) \[DifferentialD]y

4.

(∫_0^2) (1/(8 + 2  (x^2))) \[DifferentialD]x

6.

(∫_0^((1/(2 (Sqrt[ 2])))))(2/(Sqrt[1 - 4  (x^2)])) 
        \[DifferentialD]x

8.

∫(Sqrt[1 - 9  (t^2)]) \[DifferentialD]t

10.

∫(5/(Sqrt[25  (x^2) - 9])) \[DifferentialD]x

12.

∫((Sqrt[(y^2) - 25])/(y^3)) \[DifferentialD]y

16.

∫(1/((x^2) (Sqrt[(x^2) + 1]))) \[DifferentialD]x

20.

(∫_0^1) (1/(((4 - (x^2)))^(3/2))) \[DifferentialD]x

22.

∫((x^2)/((((x^2) - 1))^(5/2))) \[DifferentialD]x

Simplify[%]

24.

∫((Sqrt[1 - (x^2)])/(x^4)) \[DifferentialD]x

Simplify[%]

After trig substitution x  =  cos(θ), \(TraditionalForm
      \`dx\  = \ 
        \(-\(sin(θ)\)\) dθ\), (Sqrt[1 - (x^2)]) = sin(θ), the integral becomes \(TraditionalForm
      \`\(-\(∫\(\(\(sin\^2\) \((θ)\)\)\/\(\(cos\^4\)(
                    θ)\)\) dθ\)\)\  = \ 
        \(-\(∫\(\(tan\^2\)(θ)\) \(\(sec\^2\)(θ)\) 
              dθ\)\)\). Using the substitution u  =  tan(θ), \(TraditionalForm
      \`du = \ 
        \(\(sec\^2\)(θ)\) dθ\) yields \(TraditionalForm
      \`\(-\(∫\(u\^2\) du\)\)\  = 
        \(\(-\ \(\(\ 1\)\/3\)\) u\^3 + \ C\  = 
          \(\(-\ \(1\/3\)\) \(\(tan\^3\)(θ)\)\  + C\  = 
            \(\(-\ \(1\/3\)\) \((\@\(1 - x\^2\)\/x)\)\^3 + C = 
              \(-\(1\/3\)\) \((1 - x\^2)\)\^\(3/2\)\/x\^3 + C\)\)\)\)

30.

∫((E^t)/(((1 + (E^(2  t))))^(3/2))) 
        \[DifferentialD]t

(∫_(Log[3/4])^(Log[4/3]
          ))((E^t)/(((1 + (E^(2  t))))^(3/2))) \[DifferentialD]t

Substitute u  =  (e^t), \(TraditionalForm
      \`du\  = \ \(e\^t\) dt\) to get \(TraditionalForm
      \`∫\(u\/\((1 + u\^2)\)\^\(3/2\)\) du\). Now use a trig substitution, u  =  tan(θ), \(TraditionalForm
      \`du\  = \ 
        \(\(sec\^2\)(θ)\) dθ\), (Sqrt[1 + (u^2)]) = sec(θ): \(TraditionalForm
      \`∫\(\(\(sec\^2\)(θ)\)\/\(\(sec\^3\)(θ)\)\) 
            dθ\  = \ 
        \(∫\(cos(θ)\) dθ\  = \ 
          \(sin(θ) + C\  = \ 
            \(u\/\@\(1 + u\^2\) + C\  = \ 
              e\^t\/\@\(1 + e\^\(2  t\)\) + C\)\)\)\). Plugging in the limits ln(4/3) and ln(3/4) gives \(TraditionalForm
      \`\(4/3\)\/\@\(1 + 16\/9\) - \ \(3/4\)\/\@\(1 + 9\/16\) = \ 
        \(4\/\@\(9 + 16\) - 3\/\@\(16 + 9\) = \(4\/5 - 3\/5 = 1\/5\)\)\).