Let I=∫_0^∞ log (1+x^2)/1+x^2dx Put x=tan θ⇒ dx=sec^2 θ d θ, ∴ I= ∫_0^n/2 log (sec θ)^2d θ = 2 ∫_0^n/2 log sec θ d θ = 2 ∫_0^n/2log cos θ d θ = 2. π/2 log 1/2= ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Mathematics

Property 1

∫0∞ log 1+x2/...

Question

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

$π l o g \frac{1}{2}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

π l o g 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 π l o g \frac{1}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 π l o g 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $π l o g 2$
$L e t I = \int_{0}^{\infty} \frac{l o g (1 + x^{2})}{1 + x^{2}} d x$
$P u t x = t a n θ \Rightarrow d x = s e c^{2} θ d θ,$
$∴ I = \int_{0}^{\frac{n}{2}} l o g (s e c θ)^{2} d θ = 2 \int_{0}^{\frac{n}{2}} l o g s e c θ d θ$
$= - 2 \int_{0}^{\frac{n}{2}} l o g c o s θ d θ$ $= - 2. \frac{π}{2} l o g \frac{1}{2} = - π l o g \frac{1}{2} = π l o g 2$

Suggest Corrections

Similar questions

\int_{0}^{\infty} \frac{log (1 + x^{2})}{1 + x^{2}} d x =

Q. Prove that

\int_{0}^{\infty} \frac{log (1 + x^{2})}{1 + x^{2}} d x = π log 2

Q. If

\int_{0}^{\infty} \frac{log (1 + x^{2})}{1 + x^{2}} d x = λ \int_{0}^{1} \frac{log (1 + x)}{1 + x^{2}} d x

then

λ

equals

\int {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) d x = f (x) - log (1 + x^{2}) + c

then

f (x) =

Q. Let

f (x) = \int \frac{x^{2}}{(1 + x^{2}) (1 + \sqrt{1 + x^{2}})} d x

and

f (0) = 0

, then the value of

f (1)

will be

Join BYJU'S Learning Program

∫∞0 log(1+x2)1+x2dx=

The correct option is B π log 2Let I=∫∞0 log(1+x2)1+x2dx Put x=tan θ⇒dx=sec2 θ dθ, ∴I=∫n20 log (sec θ)2dθ=2∫n20 log sec θ dθ =−2∫n20log cos θ dθ =−2. π2log12=−π log 12=π log 2

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