Chapter 7 - Principles Of Integral Evaluation - 7.2 Integration By Parts - Exercises Set 7.2 - Page 499: 66

$$\eqalign{ & \left( {\text{a}} \right)\frac{1}{3}{x^2}{e^{3x}} - \frac{2}{9}x{e^{3x}} + \frac{2}{{27}}{e^{3x}} + C \cr & \left( {\text{b}} \right)12 - \frac{{32}}{e} \cr} $$

$$\eqalign{ & \left( {\text{a}} \right)\int {{x^2}{e^{3x}}dx} \cr & {\text{Let }}u = 3x,{\text{ }}x = \frac{u}{3},{\text{ }}dx = \frac{1}{3}du \cr & \int {{x^2}{e^{3x}}dx} = \int {{{\left( {\frac{u}{3}} \right)}^2}{e^u}\left( {\frac{1}{3}} \right)du} \cr & {\text{ }} = \frac{1}{{27}}\int {{u^2}{e^u}du} \cr & {\text{Use the reduction formula }}\int {{x^n}{e^x}dx = {x^n}{e^x} - n\int {{x^{n - 1}}{e^x}dx} } \cr & {\text{ }} = \frac{1}{{27}}\left[ {{u^2}{e^u} - 2\int {u{e^u}du} } \right] \cr & {\text{ }} = \frac{1}{{27}}{u^2}{e^u} - \frac{2}{{27}}\int {u{e^u}du} \cr & {\text{ }} = \frac{1}{{27}}{u^2}{e^u} - \frac{2}{{27}}\left( {u{e^u} - \int {{e^u}du} } \right) \cr & {\text{ }} = \frac{1}{{27}}{u^2}{e^u} - \frac{2}{{27}}\left( {u{e^u} - {e^u}} \right) + C \cr & {\text{ }} = \frac{1}{{27}}{u^2}{e^u} - \frac{2}{{27}}u{e^u} + \frac{2}{{27}}{e^u} + C \cr & {\text{Substitute - back }}u = 3x \cr & {\text{ }} = \frac{1}{{27}}{\left( {3x} \right)^2}{e^{3x}} - \frac{2}{{27}}\left( {3x} \right){e^{3x}} + \frac{2}{{27}}{e^{3x}} + C \cr & {\text{ }} = \frac{1}{3}{x^2}{e^{3x}} - \frac{2}{9}x{e^{3x}} + \frac{2}{{27}}{e^{3x}} + C \cr & \left( {\text{b}} \right)\int_0^1 {x{e^{ - \sqrt x }}dx} \cr & {\text{Let }}u = - \sqrt x ,{\text{ }}x = {u^2},{\text{ }}dx = 2udu \cr & {\text{The new limits are}} \cr & x = 1 \Rightarrow u = - 1 \cr & x = 0 \Rightarrow u = 0 \cr & \int_0^1 {x{e^{ - \sqrt x }}dx} = 2\int_0^{ - 1} {{u^3}{e^u}du} \cr & \cr & {\text{Use the reduction formula }}\int {{x^n}{e^x}dx = {x^n}{e^x} - n\int {{x^{n - 1}}{e^x}dx} } \cr & {\text{ }} = 2\left[ {{u^3}{e^u} - 3\int {{u^2}{e^u}du} } \right] \cr & {\text{ }} = 2{u^3}{e^u} - 6\left[ {{u^2}{e^u} - 2\int {u{e^u}du} } \right] \cr & {\text{ }} = 2{u^3}{e^u} - 6{u^2}{e^u} + 12\int {u{e^u}du} \cr & {\text{ }} = 2{u^3}{e^u} - 6{u^2}{e^u} + 12\left[ {u{e^u} - \int {{e^u}du} } \right] \cr & {\text{ }} = 2{u^3}{e^u} - 6{u^2}{e^u} + 12u{e^u} - 12{e^u} + C \cr & {\text{Therefore}} \cr & 2\int_0^{ - 1} {{u^3}{e^u}du} = \left[ {2{u^3}{e^u} - 6{u^2}{e^u} + 12u{e^u} - 12{e^u}} \right]_0^{ - 1} \cr & {\text{ }} = \left[ {2{{\left( { - 1} \right)}^3}{e^{ - 1}} - 6{{\left( { - 1} \right)}^2}{e^{ - 1}} + 12\left( { - 1} \right){e^{\left( { - 1} \right)}} - 12{e^{ - 1}}} \right] + \left[ {12{e^0}} \right] \cr & {\text{Simplifying}} \cr & {\text{ }} = - \frac{2}{e} - \frac{6}{e} - \frac{{12}}{e} - \frac{{12}}{e} + 12 \cr & {\text{ }} = 12 - \frac{{32}}{e} \cr} $$