∫_1 ^3 x^(2x) (1+ln x) dx =?


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Question Number 125962 by bramlexs22 last updated on 15/Dec/20

$\underset{1}{\int^{3}} x^{2 x} (1 + \ln x) d x = ?$
	Answered by liberty last updated on 16/Dec/20
	$let x^x = r → { ((x=1→r=1)),((x=3→r=27)) :} ∧ dr = x^x (1+ln x)dx I=∫_1 ^3 x^x .x^x (1+ln x) dx = ∫_1 ^(27) r dr I= (1/2)[ r^2 ]_1 ^(27) = (1/2)×28×26 = 14×26= 364$
	$l e t x^{x} = r \to {\begin{cases} x = 1 \to r = 1 \\ x = 3 \to r = 27 \end{cases} \land d r = x^{x} (1 + \ln x) d x$ $I = \underset{1}{\int^{3}} x^{x} . x^{x} (1 + \ln x) d x = \underset{1}{\int^{27}} r d r$ $I = \frac{1}{2} {[r^{2}]}_{1}^{27} = \frac{1}{2} \times 28 \times 26 = 14 \times 26 = 364$
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