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Question Number 5422 by love math last updated on 14/May/16

∫_0 ^(7/2) (x+1)(2x+1)^(1/3) dx

$$\int_{\mathrm{0}} ^{\frac{\mathrm{7}}{\mathrm{2}}} \left({x}+\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {dx} \\ $$

Answered by nchejane last updated on 14/May/16

=[(3/8)(x+1)(2x+1)^(4/3) ]_0 ^(7/2) −(3/8)∫_0 ^(7/2) (2x+1)^(4/3) dx  =[(3/8)(x+1)(2x+2)^(4/3) ]_0 ^(7/2) −[(9/(28))(2x+1)^(7/3) ]_0 ^(7/2)   =....

$$=\left[\frac{\mathrm{3}}{\mathrm{8}}\left({x}+\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} \right]_{\mathrm{0}} ^{\frac{\mathrm{7}}{\mathrm{2}}} −\frac{\mathrm{3}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\mathrm{7}}{\mathrm{2}}} \left(\mathrm{2}{x}+\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} {dx} \\ $$$$=\left[\frac{\mathrm{3}}{\mathrm{8}}\left({x}+\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{2}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} \right]_{\mathrm{0}} ^{\frac{\mathrm{7}}{\mathrm{2}}} −\left[\frac{\mathrm{9}}{\mathrm{28}}\left(\mathrm{2}{x}+\mathrm{1}\right)^{\frac{\mathrm{7}}{\mathrm{3}}} \right]_{\mathrm{0}} ^{\frac{\mathrm{7}}{\mathrm{2}}} \\ $$$$=.... \\ $$

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