Solveeit Logo
Login

Question

Question: The value of integral \(\int_{0}^{\infty}\frac{x\log x}{(1 + x^{2})^{2}}\)dx is –...

The value of integral ∫0∞xlog⁡x(1+x2)2\int_{0}^{\infty}\frac{x\log x}{(1 + x^{2})^{2}}dx is –

A

1

B

0

C

2

D

None of these

Answer

0

Explanation

Solution

I = ∫01xlog⁡x(1+x2)2\int_{0}^{1}\frac{x\log x}{(1 + x^{2})^{2}}dx + ∫1∞xlog⁡x(1+x2)2\int_{1}^{\infty}\frac{x\log x}{(1 + x^{2})^{2}}dx

= I1 + I2.

Put x = 1t\frac{1}{t} in I2 and adjust the limits

I2 = ∫101tlog⁡1t.(−1t2)dt(1+1t2)2\int_{1}^{0}\frac{\frac{1}{t}\log\frac{1}{t}.\left( \frac{- 1}{t^{2}} \right)dt}{\left( 1 + \frac{1}{t^{2}} \right)^{2}}= ∫01−tlog⁡t(1+t2)2\int_{0}^{1}\frac{- t\log t}{(1 + t^{2})^{2}}dt

I­2 = – ∫01xlog⁡x(1+x2)2\int_{0}^{1}\frac{x\log x}{(1 + x^{2})^{2}}dx = –I1

Hence I1 + I2 = 0.