7.Rev, #20. Use formula 25 on page 367 to get

∫ \frac{5x + 6}{x^{2} + 4} dx = \frac{5}{2} ∫ \frac{2x}{x^{2} + 4} dx + 6 ∫ \frac{dx}{x^{2} + 4} = \frac{5}{2} ln(x^{2} + 4) + 3 arctan \frac{x}{2} + C.

7.Rev, #25. ∫ \frac{(u + 1)^{3}}{u^{2}} du = ∫ \frac{u^{3} + 3u^{2} + 3u + 1}{u^{2}} du = \frac{1}{2}u^{2} + 3u + 3 ln |u| − \frac{1}{u} + C.

7.Rev, #32. u = x^{2}, du = 2x dx changes ∫ \frac{x dx}{\sqrt{1 − x^{4}}} to \frac{1}{2} ∫ \frac{du}{\sqrt{1 − u^{2}}} .

7.Rev, #35.

∫ \frac{e^{x}}{1 + e^{x}} dx = ln(1 + e^{x}) + C

∫ \frac{e^{x}}{1 + e^{2x}} dx = ∫ \frac{dw}{1 + w^{2}} = arctan w + C = arctan e^{x} + C

(substituting w = e^{x})

∫ \frac{dx}{1 + e^{x}} = ∫ \frac{dw}{w(1 + w)} = ln |w| − ln |1 + w| + C = x − ln(1 + e^{x}) + C

(substituting w = e^{x} and using formula 26 on page 367)

7.Rev, #36. The averages over one cycle in ascending order are

\frac{1}{2π} ∫_{0}^{2π} cos t dt = 0,

\frac{1}{2π} ∫_{0}^{2π} cos^{2} t dt = \frac{1}{2π} \frac{1}{2}(cos x sin x + x)|_{0}^{2π} = \frac{1}{2}

\frac{1}{2π} ∫_{0}^{2π} | cos t| dt = 4 × \frac{1}{2π} ∫_{0}^{π/2} cos t dt (by symmetry)

= \frac{2}{π} sin t|_{0}^{π/2} = \frac{2}{π}

The first average is clearly zero becuase of the oscillation of cos x. The other averages are positive since cos^{2} t and | cos t| are non-negative. Moreover, cos^{2} t = | cos t|^{2} ≤ | cos t| since | cos t| ≤ 1. So the average for cos^{2} t is less than that for | cos t|.

7.Rev, #40. y = x^{2} and y = 6 − x intersect at (2, 4). Since a vertical strip of the shaded region lying over x has length 6 − x − x^{2}, the average vertical height of the shaded region is

\frac{1}{2} ∫_{0}^{2}(6 − x − x^{2}) dx = \frac{1}{2}(6x − \frac{x^{2}}{2} − \frac{x^{3}}{3} )|_{0}^{2} = \frac{1}{2}(12 − 2 − (8/3)) = 11/3

7.Rev, #51. The integral is improper since 1/\sqrt{t + 5} is undefined at t = −5. The integral converges since

lim_{b→−5^{+}} ∫_{b}^{10} \frac{dt}{\sqrt{t + 5}} = lim_{b→−5^{+}} 2\sqrt{t + 5}|_{b}^{10} = lim_{b→−5^{+}} 2(\sqrt{15} − \sqrt{b + 5}) = 2\sqrt{15}.