\[\int\limits_0^1 \frac{2x}{1 + x^2} dx\]

\[\text{We have}, \]\[I = \int_0^1 \frac{2x}{1 + x^2} d x\]\[\text{Putting} 1 + x^2 = t\]\[ \Rightarrow 2x\ dx = dt\]\[\text{When } x \to 0; t \to 1\]\[\text{And } x \to 1; t \to 2\]\[ \therefore I = \int_1^2 \frac{d t}{t}\]\[ = \left[ \log_e \left| t \right| \right]_1^2 \]\[ = \log_e 2 - \log_e 1\]\[ = \log_e 2 - 0\]\[ = \log_e 2\]

1 ∫ 0 2 X 1 + X 2 D X - Mathematics

प्रश्न

\int_{0}^{1} \frac{2 x}{1 + x^{2}} d x

बेरीज

उत्तर

$We have,$
$I = \int_{0}^{1} \frac{2 x}{1 + x^{2}} d x$
$Putting 1 + x^{2} = t$
$\Rightarrow 2 x d x = d t$
$When x \to 0; t \to 1$
$And x \to 1; t \to 2$
$∴ I = \int_{1}^{2} \frac{d t}{t}$
$= {[\log_{e} | t |]}_{1}^{2}$
$= \log_{e} 2 - \log_{e} 1$
$= \log_{e} 2 - 0$
$= \log_{e} 2$

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Definite Integrals

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Q 25 Q 24 Q 26

पाठ 20: Definite Integrals - Very Short Answers [पृष्ठ ११५]

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पाठ 20 Definite Integrals
Very Short Answers | Q 25 | पृष्ठ ११५

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