find lim_(x→0) ∫_(x+1) ^(2x+1) (t^2 /(ln(1+t)))dt .


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Question Number 28429 by abdo imad last updated on 25/Jan/18

$f i n d {l i m}_{x \to 0} \int_{x + 1}^{2 x + 1} \frac{t^{2}}{l n (1 + t)} d t .$
Commented by abdo imad last updated on 27/Jan/18

$\exists c \in] x + 1, 2 x + 1 [/ \int_{x + 1}^{2 x + 1} = \frac{1}{l n (1 + c)} \int_{x + 1}^{2 x + 1} t^{2} d t$ $= \frac{1}{3} \frac{1}{l n (1 + c)} ({(2 x + 1)}^{3} - {(x + 1)}^{3}) w e h a v e x \to 0 \Rightarrow c \to 1$ $a n d {l i m}_{x \to 0} \int_{x + 1}^{2 x + 1} \frac{t^{2}}{l n (1 + t)} d t = \frac{1}{3 l n 2} \times 0 = 0$ $(l o o k t h a t t h e f u n c t i o n w (t) = \frac{1}{l n (1 + t)} i s c o n t i n u e o n$ $[x + 1, 2 x + 1] .)$
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