The correct option is A π4ln2 ⇒ i( ln(1 i))ln(1x+il)+ln(i+1)ln(1x il)+li2(x+ix 1)li2( x+i1+i))+c2 ∴ arctan(x)ln(x+1) arctan|(x)ln(x^2+2x+12) tan ...

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Integration Using Substitution

∫01ln1+x1+x2d...

Question

$\int_{0}^{1} \frac{l n (1 + x)}{1 + x^{2}} d x =$

\frac{π}{4} l n 2

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\frac{π}{2} l n 2

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\frac{π}{8} l n 2

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π l n 2

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Solution

The correct option is A $\frac{π}{4} l n 2$
$\Rightarrow \frac{- i (- l n (1 - i)) l n (1 x + i l) + l n (i + 1) l n (1 x - i l) + l i 2 (\frac{x + i}{x - 1}) l i 2 (\frac{- x + i}{1 + i})) + c}{2}$
$∴ a r c tan (x) l n (x + 1) - a r c tan | (x) l n (\frac{x^{2} + 2 x + 1}{2}) - t a n 2 (\frac{x + 1}{2}, \frac{x + 1}{2})$
$\frac{l n (x^{2} + 1) - i l i 2 (l \frac{l (i + 1) x - i + 1}{2}) + i l i 2 (\frac{i (i + 1) x - i - 1}{2})^{+ c}}{2}$

$\frac{∴ - i l i 2 (\frac{i + 1}{2}) - i l i 2 (\frac{- i - 1}{2}) + \frac{i l i 2 (- 1)}{2} - \frac{- i (l i 2 - 1)}{2} + \frac{π l n (2)}{4}}{2}$
$= \frac{π l n (2)}{4}$

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