(1)
let
x^5 =√2 tanu
5x^4 dx =√2 (secu)^2 du
∫ dx/[x(2+x)^10]
=(1/5) ∫ 5x^4dx/[x^5.(2+x^10) ]
=(1/5) ∫ √2 (secu)^2 du/[√2 tanu . (2secu) ]
=(1/10) ∫ cscu du
=(1/10)ln|cscu -cotu| + C
=(1/10)ln|√(2-x^10)/x^5 -√2/x^5| + C
=(1/10)ln|√(2-x^10) -√2| - (1/2)ln|x| + C
(2)
∫ xsinx.cosx dx
=(1/2) ∫ xsin2x dx
=-(1/4) ∫ x dcos2x
=-(1/4) xcos2x + (1/4) ∫ cos2x dx
=-(1/4) xcos2x + (1/8) sin2x + C
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