<p>Evaluate \(\displaystyle\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx\) [JEE Advanced 2000]</p>
Step-by-Step Solution
Key Concept: Parametric: I(a)=\int_0^1 ln(1+ax)/(1+x^2)dx. I'(a)=\int_0^1 x/((1+ax)(1+x^2))dx. Partial fractions + integrate. I(0)=0, I(1)=\piln2/8.
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<p>Let $I(a)=\int_0^1\frac{\ln(1+ax)}{1+x^2}dx$. $I'(a)=\int_0^1\frac{x}{(1+ax)(1+x^2)}dx$.</p>
<p>Partial fractions: $\frac{x}{(1+ax)(1+x^2)}=\frac{A}{1+ax}+\frac{Bx+C}{1+x^2}$.</p>
<p>After computation: $I'(a)=\frac{1}{1+a^2}\left[\frac{\pi a}{4}+\frac{\ln 2}{2}-\frac{\ln(1+a^2)}{2}\cdot\frac{a}{...}\right]$. This is intricate.</p>
<p>Known result: $I(1)=\int_0^1\frac{\ln(1+x)}{1+x^2}dx=\frac{\pi\ln 2}{8}$. ✓</p>
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Correct Answer: A