The correct option is C 0I=∫_0^∞log x/1+x^2dx I=∫_0^∞log x/1+x^2dx Let x=1/t I=∫_0^∞log1/t/1+1/t^2 ( dt/t^2 ) ⇒ 2I=∫_0^∞log x/1+x^2dx ∫_0^∞log t/1+t^2dt=0 ∴ ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

∫0∞log x/1+x2...

Question

$\int_{0}^{\infty} \frac{l o g x}{1 + x^{2}} d x =$

\frac{π}{4}

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\frac{π}{4} l o g \frac{π}{2}

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T a n^{- 1} (\frac{1}{\sqrt{2}})

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Solution

The correct option is D $0$
$I = \int_{0}^{\infty} \frac{l o g x}{1 + x^{2}} d x$
$I = \int_{0}^{\infty} \frac{l o g x}{1 + x^{2}} d x$
Let $x = \frac{1}{t}$
$I = \int_{0}^{\infty} \frac{l o g \frac{1}{t}}{1 + \frac{1}{t^{2}}} (- \frac{d t}{t^{2}})$
$\Rightarrow 2 I = \int_{0}^{\infty} \frac{l o g x}{1 + x^{2}} d x - \int_{0}^{\infty} \frac{l o g t}{1 + t^{2}} d t = 0$
$∴ I = 0$

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∫∞0log x1+x2dx=

The correct option is D 0I=∫∞0log x1+x2dx I=∫∞0log x1+x2dx Let x=1t I=∫∞0log1t1+1t2(−dtt2) ⇒2I=∫∞0log x1+x2dx−∫∞0log t1+t2dt=0 ∴ I=0

$\int_{0}^{\infty} \frac{l o g x}{1 + x^{2}} d x =$