1. 3^-1/2 ln |x + √ (2/3 + x²)| + C

    1. The closest formula is number 20. from the notes, except for the “3” in front of x². We can            pull it out of the parentheses: [2 + 3x²]^-1/2= 3^-1/2 [2/3 + x²]^-1/2 Now we can use the formula,            with a = √ (2/3): ∫[2 + 3x²]^-1/2 dx = ∫ 3^-1/2 [2/3 + x²>]^-1/2 dx = 3^-1/2 ln | x + √ (2/3 + x²) | + C

    1. -1/3 sin²(x) cos(x) – 2/3 cos(x) + c

     1. We can use formula 7. from the notes, with n = 2: ∫sin³(x) dx = – 1/3 sin²(x) cos(x) + 2/3 ∫                   sin(x) dx = – 1/3 sin²(x) cos(x) – 2/3 cos(x) + c

1. 1/2 tan (2x) – x + c

    1. The notes have formula 3. for tan²(x). To use it, first substitute u = 2x, so du = 2 dx, and we            get: ∫tan²(2x) dx = ∫ 1/2 tan²(u) du = 1/2 [ tan(u) – u ] + c= 1/2 tan (2x) -x + c

    1. x² arccos(x²)- √(1 – x⁴) + C

    1. Formula 47. in the notes has arccos (or cos^-1) in it. We can’t use it as stated, since we need x² inside the arccos, and we have an extra 2x up front. This is an ideal case for substitution, since 2x is the derivative of x². Use u = x², so du = 2x dx: ∫2x arccos(x²) dx = ∫ arccos(u) du. Now we use formula 47.: int; arccos(u) du = u arccos(u) – √ ( 1 – u2 ) + C Substituting back x² for u, we get the answer: ∫ 2x arccos(x²) dx = x² arccos(x²) – √(1 – x⁴) + C

    1. Arctan(x + 1) + c

1. This does not look like anything in the tables. We have formulas for things such as a² + x², but      not summand of “x” in there. We have to put our expression into that form first, by absorbing the “2x” into the square: x² + 2x + 2 = (x² + 2x + 1) + 1 = (x + 1)² + 1 With this, we have: ∫[x² + 2x + 2]^-1 dx = ∫[( x + 1)² + 1]^-1 dx This is close to formula 41. in the notes, if we substitute u = x + 1, du = dx: ∫[x² + 2x + 2]^-1 dx = ∫[(x + 1)2 + 1]^-1 dx = ∫[u² + 1]^-1 du = arctan(u) + c = arctan(x + 1) + c