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Lösung 2.2:2d

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What makes the integral not entirely simple is the expression 1x under the root sign, so we try the substitution u=1x,

1031xdx=udu=1x=(1x)dx=dx=013udu. 

Note how the new limits of integration go from 1 to 0 (and not the other way around!). It is possible to change the order of the limits if we change sign at the same time, i.e.

013udu=+103udu. 

All that is now left is routine calculations,

103udu=10u13du= 31+1u13+1 10=43 u43 10=43143043=43.