Question Number 159915 by tounghoungko last updated on 22/Nov/21
$$\:\:\int_{\mathrm{1}} ^{\mathrm{16}} \:\frac{\sqrt{{x}}}{\mathrm{1}+\sqrt[{\mathrm{4}}]{{x}^{\mathrm{3}} }}\:{dx}\:=? \\ $$
Commented by blackmamba last updated on 22/Nov/21
![let (x)^(1/4) = v ⇒ { ((v_1 =1)),((v_2 =2)) :} ⇒x=v^4 ; dx=4v^3 dv M=∫_1 ^( 2) (v^2 /(1+v^3 )) (4v^3 ) dv M= 4∫_1 ^( 2) ((v^2 (v^3 +1)−v^2 )/(1+v^3 )) dv M=4 [∫_1 ^( 2) v^2 dv−∫_1 ^( 2) (v^2 /(1+v^3 )) dv ] M=4 [(1/3)v^3 ]_1 ^2 −(4/3)∫_1 ^( 2) ((d(1+v^3 ))/(1+v^3 )) M=((28)/3) −(4/3)[ ln (1+v^3 ) ]_1 ^2 M= ((28)/3)−(4/3)(ln 9−ln 2) M=(4/3) (7−ln 9 +ln 2)](https://www.tinkutara.com/question/Q159922.png)
$$\:\:\:{let}\:\sqrt[{\mathrm{4}}]{{x}}\:=\:{v}\:\Rightarrow\begin{cases}{{v}_{\mathrm{1}} =\mathrm{1}}\\{{v}_{\mathrm{2}} =\mathrm{2}}\end{cases}\:\Rightarrow{x}={v}^{\mathrm{4}} \:;\:{dx}=\mathrm{4}{v}^{\mathrm{3}} \:{dv} \\ $$$$\:\:\:{M}=\int_{\mathrm{1}} ^{\:\mathrm{2}} \:\frac{{v}^{\mathrm{2}} }{\mathrm{1}+{v}^{\mathrm{3}} }\:\left(\mathrm{4}{v}^{\mathrm{3}} \right)\:{dv}\: \\ $$$$\:\:\:{M}=\:\mathrm{4}\int_{\mathrm{1}} ^{\:\mathrm{2}} \:\frac{{v}^{\mathrm{2}} \left({v}^{\mathrm{3}} +\mathrm{1}\right)−{v}^{\mathrm{2}} }{\mathrm{1}+{v}^{\mathrm{3}} }\:{dv} \\ $$$$\:\:{M}=\mathrm{4}\:\left[\int_{\mathrm{1}} ^{\:\mathrm{2}} {v}^{\mathrm{2}} \:{dv}−\int_{\mathrm{1}} ^{\:\mathrm{2}} \frac{{v}^{\mathrm{2}} }{\mathrm{1}+{v}^{\mathrm{3}} }\:{dv}\:\right] \\ $$$$\:\:{M}=\mathrm{4}\:\left[\frac{\mathrm{1}}{\mathrm{3}}{v}^{\mathrm{3}} \:\right]_{\mathrm{1}} ^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{1}} ^{\:\mathrm{2}} \:\frac{{d}\left(\mathrm{1}+{v}^{\mathrm{3}} \right)}{\mathrm{1}+{v}^{\mathrm{3}} }\: \\ $$$$\:\:{M}=\frac{\mathrm{28}}{\mathrm{3}}\:−\frac{\mathrm{4}}{\mathrm{3}}\left[\:\mathrm{ln}\:\left(\mathrm{1}+{v}^{\mathrm{3}} \right)\:\right]_{\mathrm{1}} ^{\mathrm{2}} \: \\ $$$$\:\:\:{M}=\:\frac{\mathrm{28}}{\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{ln}\:\mathrm{9}−\mathrm{ln}\:\mathrm{2}\right) \\ $$$$\:\:\:{M}=\frac{\mathrm{4}}{\mathrm{3}}\:\left(\mathrm{7}−\mathrm{ln}\:\mathrm{9}\:+\mathrm{ln}\:\mathrm{2}\right)\: \\ $$