Chapter 7 - Principles Of Integral Evaluation - 7.2 Integration By Parts - Exercises Set 7.2 - Page 499: 54

$$\eqalign{ & \left( {\text{a}} \right)\frac{{2 - \sqrt 2 }}{3} \cr & \left( {\text{b}} \right)\frac{{2 - \sqrt 2 }}{3} \cr} $$

$$\eqalign{ & \int_0^1 {\frac{{{x^3}}}{{\sqrt {{x^2} + 1} }}} dx \cr & \left( {\text{a}} \right) \cr & \int {\frac{{{x^3}}}{{\sqrt {{x^2} + 1} }}dx = \int {\left( {{x^2}} \right)\left( {\frac{x}{{\sqrt {{x^2} + 1} }}} \right)} } dx \cr & {\text{Let }}u = {x^2},{\text{ }}du = 2xdx \cr & dv = \frac{x}{{\sqrt {{x^2} + 1} }}dx,{\text{ }}v = \sqrt {{x^2} + 1} \cr & {\text{Using the integration by parts formula}} \cr & \int {\frac{{{x^3}}}{{\sqrt {{x^2} + 1} }}dx} = {x^2}\sqrt {{x^2} + 1} - \int {\sqrt {{x^2} + 1} \left( {2x} \right)dx} \cr & {\text{ }} = {x^2}\sqrt {{x^2} + 1} - \frac{{{{\left( {{x^2} + 1} \right)}^{3/2}}}}{{3/2}} + C \cr & {\text{ }} = {x^2}\sqrt {{x^2} + 1} - \frac{{2{{\left( {{x^2} + 1} \right)}^{3/2}}}}{3} + C \cr & {\text{Therefore,}} \cr & \int_0^1 {\frac{{{x^3}}}{{\sqrt {{x^2} + 1} }}} dx = \left[ {{x^2}\sqrt {{x^2} + 1} - \frac{{2{{\left( {{x^2} + 1} \right)}^{3/2}}}}{3}} \right]_0^1 \cr & = \left[ {{1^2}\sqrt {{1^2} + 1} - \frac{{2{{\left( {{1^2} + 1} \right)}^{3/2}}}}{3}} \right] - \left[ {{0^2}\sqrt {{0^2} + 1} - \frac{{2{{\left( {{0^2} + 1} \right)}^{3/2}}}}{3}} \right] \cr & = \sqrt 2 - \frac{2}{3}\left( {2\sqrt 2 } \right) + \frac{2}{3} \cr & = \frac{2}{3} - \frac{1}{3}\sqrt 2 \cr & = \frac{{2 - \sqrt 2 }}{3} \cr & \cr & \left( {\text{b}} \right) \cr & \int {\frac{{{x^3}}}{{\sqrt {{x^2} + 1} }}dx} \cr & {\text{Let }}u = \sqrt {{x^2} + 1} ,{\text{ }}du = \frac{x}{{\sqrt {{x^2} + 1} }}dx \cr & \int {\frac{{{x^3}}}{{\sqrt {{x^2} + 1} }}dx} = \int {{x^2}du} \cr & u = \sqrt {{x^2} + 1} \Rightarrow {x^2} = {u^2} - 1 \cr & \int {{x^2}du} = \int {\left( {{u^2} - 1} \right)du} \cr & {\text{Integrating}} \cr & {\text{ = }}\frac{1}{3}{u^3} - u + C \cr & {\text{Back - substitute }}u = \sqrt {{x^2} + 1} \cr & {\text{ = }}\frac{1}{3}{\left( {\sqrt {{x^2} + 1} } \right)^3} - \sqrt {{x^2} + 1} + C \cr & \int_0^1 {\frac{{{x^3}}}{{\sqrt {{x^2} + 1} }}} dx = \left[ {\frac{1}{3}{{\left( {\sqrt {{x^2} + 1} } \right)}^3} - \sqrt {{x^2} + 1} } \right]_0^1 \cr & = \left[ {\frac{1}{3}{{\left( {\sqrt {{1^2} + 1} } \right)}^3} - \sqrt {{1^2} + 1} } \right] - \left[ {\frac{1}{3}{{\left( {\sqrt {{0^2} + 1} } \right)}^3} - \sqrt {{0^2} + 1} } \right] \cr & = \frac{1}{3}\left( {2\sqrt 2 } \right) - \sqrt 2 - \frac{1}{3} + 1 \cr & = - \frac{1}{3}\sqrt 2 + \frac{2}{3} \cr & = \frac{{2 - \sqrt 2 }}{3} \cr} $$