-01log 1-x-1-x2 d x-A- -4log 2B- -8log 2C- -log 2D- none

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

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Byju's Answer

Standard XII

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

∫01log 1+x/1+...

Question

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

\frac{π}{8} l o g 2

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\frac{π}{4} l o g 2

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π l o g 2

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n o n e

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Solution

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

Suggest Corrections

∫10 log(1+x)1+x2 dx=

The correct option is A π8 log 2If x=tan θ, I=∫10 log(1+x)1+x2 dx=∫π20 log(1+tan θ)sec2 θsec2 θ d θ=∫π40 log (1+tan θ)d θ =∫π40(1+tan(π4−θ))dθ=∫π40(21+tan θ)dθ 2I=θπ40log 2 dθ=log 2π4⇒I=(log 2)π8

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