∫_1 ^(16) ((√x)/(1+(x^3 )^(1/4) )) dx =?


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Question Number 159915 by tounghoungko last updated on 22/Nov/21

$\int_{1}^{16} \frac{\sqrt{x}}{1 + \sqrt[4]{x^{3}}} d x = ?$
Commented by blackmamba last updated on 22/Nov/21
$let (x)^(1/4) = v ⇒ { ((v_1 =1)),((v_2 =2)) :} ⇒x=v^4 ; dx=4v^3 dv M=∫_1 ^( 2) (v^2 /(1+v^3 )) (4v^3 ) dv M= 4∫_1 ^( 2) ((v^2 (v^3 +1)−v^2 )/(1+v^3 )) dv M=4 [∫_1 ^( 2) v^2 dv−∫_1 ^( 2) (v^2 /(1+v^3 )) dv ] M=4 [(1/3)v^3 ]_1 ^2 −(4/3)∫_1 ^( 2) ((d(1+v^3 ))/(1+v^3 )) M=((28)/3) −(4/3)[ ln (1+v^3 ) ]_1 ^2 M= ((28)/3)−(4/3)(ln 9−ln 2) M=(4/3) (7−ln 9 +ln 2)$
$l e t \sqrt[4]{x} = v \Rightarrow {\begin{cases} v_{1} = 1 \\ v_{2} = 2 \end{cases} \Rightarrow x = v^{4}; d x = 4 v^{3} d v$ $M = \int_{1}^{2} \frac{v^{2}}{1 + v^{3}} (4 v^{3}) d v$ $M = 4 \int_{1}^{2} \frac{v^{2} (v^{3} + 1) - v^{2}}{1 + v^{3}} d v$ $M = 4 [\int_{1}^{2} v^{2} d v - \int_{1}^{2} \frac{v^{2}}{1 + v^{3}} d v]$ $M = 4 {[\frac{1}{3} v^{3}]}_{1}^{2} - \frac{4}{3} \int_{1}^{2} \frac{d (1 + v^{3})}{1 + v^{3}}$ $M = \frac{28}{3} - \frac{4}{3} {[\ln (1 + v^{3})]}_{1}^{2}$ $M = \frac{28}{3} - \frac{4}{3} (\ln 9 - \ln 2)$ $M = \frac{4}{3} (7 - \ln 9 + \ln 2)$
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