The value of -018 log 1- x -1- x 2 dx isA- -2log 2B- log 2C- -log 2D- -8log 2

The correct option is C π log 2I=8 ∫_0^1 log (1+x)/1+x^2dx =8 ∫_0^π/4log(1+tan θ)/1+tan^2 θsec^2 θ d θ (Let x=tan θ) =8 ∫_0^π/4log ( 1+tan ( π/4 θ ) )d θ = ...

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Byju's Answer

Standard XII

Mathematics

Property 1

The value of ...

Question

The value of $\int_{0}^{1} \frac{8 l o g (1 + x)}{1 + x^{2}} d x i s$

\frac{π}{8} l o g 2

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\frac{π}{2} l o g 2

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l o g 2

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π l o g 2

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Solution

The correct option is D $π l o g 2$
$I = 8 \int_{0}^{1} \frac{l o g (1 + x)}{1 + x^{2}} d x$
$= 8 \int_{0}^{\frac{π}{4}} \frac{l o g (1 + t a n θ)}{1 + t a n^{2} θ} s e c^{2} θ d θ (L e t x = t a n θ)$
$= 8 \int_{0}^{\frac{π}{4}} l o g (1 + t a n (\frac{π}{4} - θ)) d θ$
$= 8 \int_{0}^{\frac{π}{4}} l o g (1 + \frac{1 - t a n θ}{1 + t a n θ}) d θ$
$= 8 \int_{0}^{\frac{π}{4}} l o g 2 d θ - 8 \int_{0}^{\frac{π}{4}} l o g (1 + t a n θ) d θ$
$= 8 l o g 2 \frac{π}{4} - I$
$2 I = 2 π l o g 2$
$I = π l o g 2$

Suggest Corrections

The value of ∫10 8 log (1+x)1+x2dx is

The correct option is D π log 2I=8∫10log(1+x)1+x2dx =8∫π40log(1+tan θ)1+tan2 θsec2 θ dθ (Let x=tan θ) =8∫π40log (1+tan(π4−θ))dθ =8∫π40log(1+1−tanθ1+tan θ)dθ =8∫π40log 2 dθ−8∫π40log (1+tanθ)dθ =8log2π4−I 2I=2π log2 I=πlog 2

The value of $\int_{0}^{1} \frac{8 l o g (1 + x)}{1 + x^{2}} d x i s$