JEE Advanced 2019 Paper 2 · Mathematics · Indefinite and Definite Integrals · medium

The value of (16)/(π)∫0^1(ln (1+x))/(1+x^2 dx is ______.

Numerical answer

The value of

\dfrac{16}{\pi}\displaystyle\int_0^1\dfrac{\ln(1+x)}{1+x^2}\,dx

is ______. (Give the answer correct to two decimal places.)

Show answer and solution

Answer: 1.39

Put

x=\tan\theta

, so the integral becomes

\int_0^{\pi/4}\ln(1+\tan\theta)\,d\theta=I

. Apply the King property

\theta\to\frac\pi4-\theta

: since

1+\tan\left(\frac\pi4-\theta\right)=\frac{2}{1+\tan\theta}

, we get

I=\int_0^{\pi/4}\left[\ln2-\ln(1+\tan\theta)\right]d\theta=\frac\pi4\ln2-I

. Therefore

I=\frac{\pi\ln2}{8}

. The required quantity is

\frac{16}{\pi}\cdot\frac{\pi\ln 2}{8}=2\ln2\approx1.3863

, which is

1.39

to two decimal places. A numerical integration of the original integral confirms the value

1.3863

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