7.3, #17. Set w = x^{2} + 4, dw = 2x dx. Then w = 20 if x = 4, w = 5 if x = 1, and

∫_{4}^{1} x\sqrt{x^{2} + 4} dx = \frac{1}{2} ∫_{20}^{5} w^{1/2} dw = \frac{1}{3}w^{3/2}|_{20}^{5} = \frac{1}{3}(5^{3/2} − 20^{3/2}) = −26.087.

7.3, #20. Set w = x^{2} + 4x + 5, dw = (2x + 4) dx. Then w = 1 if x = −2, w = 5 if x = 0, and

∫_{−2}^{0} \frac{2x + 4}{x^{2} + 4x + 5} dx = ∫_{1}^{5} \frac{dw}{w} = ln 5 − ln 1 = ln 5 = 1.609.

7.3, #27. x^{2} + 4x + 5 = (x^{2} + 4x + 4) + 1 = (x + 2)^{2} + 1. Set x + 2 = tan θ, dx = sec^{2} θ dθ. Then

∫ \frac{dx}{x^{2} + 4x + 5} = ∫ \frac{dx}{(x + 2)^{2} + 1} = ∫ \frac{sec^{2} θ}{tan^{2} θ + 1} dθ
= ∫ dθ = θ + C = arctan(x + 2) + C

using the trigonometric identity tan^{2} θ + 1 = sec^{2} θ.

7.3, #29. The area is ∫_{0}^{2} xe^{x^{2}} dx. We evaluate this by substitition. Let u = x^{2}. Then du = 2x dx. Thus the area is

∫_{0}^{2} xe^{x^{2}} dx = \frac{1}{2} ∫_{0}^{2} 2xe^{x^{2}} dx = \frac{1}{2} ∫_{0^{2}}^{2^{2}} e^{u} du = \frac{1}{2}(e^{4} − e^{0}) = \frac{1}{2}(e^{4} − 1) ≃ 26.7991.

7.4, #3. Set te^{5t} = uv′, where u = t, v′ = e^{5t}. Then u′ = 1, v = (1/5)e^{5t}, and

∫ te^{5t} dt = \frac{1}{5}te^{5t} − \frac{1}{5} ∫ e^{5t} dt = \frac{1}{5}te^{5t} − \frac{1}{25}e^{5t} + C.

7.4, #4. Set t^{2}e^{5t} = uv′, where u = t^{2}, v′ = e^{5t}. Then u′ = 2t, v = (1/5)e^{5t}, and

\begin{aligned}∫ t^{2}e^{5t} dt = \frac{1}{5}t^{2}e^{5t} − \frac{2}{5} ∫ te^{5t} dt & = \frac{1}{5}t^{2}e^{5t} − \frac{2}{5}[\frac{1}{5}te^{5t} − \frac{1}{25}e^{5t}] + C \\ & = \frac{1}{5}t^{2}e^{5t} − \frac{2}{25}te^{5t} + \frac{2}{125}e^{5t} + C \end{aligned}

7.4, #6. Set y ln y = uv′, where u = ln y, v′ = y. Then u′ = 1/y, v = y^{2}/2, and

∫ y ln y dy = \frac{1}{2}y^{2} ln y − \frac{1}{2} ∫ \frac{y^{2}}{y} dy = \frac{1}{2}y^{2} ln y − \frac{1}{4}y^{2} + C.

7.4, #7. Set x^{3} ln x = uv′, where u = ln x, v′ = x^{3}. Then u′ = 1/x, v = x^{4}/4, and

∫ x^{3} ln x dx = \frac{x^{4} ln x}{4} − \frac{1}{4} ∫ \frac{x^{4}}{x} dx = \frac{x^{4} ln x}{4} − \frac{x^{4}}{16} + C.