int_{-1}^1 frac{log (1+x)}{1+x^2} d x = int_0 | vedclass

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(C) Given the integral $I = \int_{-1}^1 \frac{\log (1+x)}{1+x^2} d x$.Split the integral at $x=0$:$I = \int_{-1}^0 \frac{\log (1+x)}{1+x^2} d x + \int

Vedclass · Accepted Answer

$\frac{\log (1-x)}{1+x^2}$

Vedclass · Answer

(C) Given the integral $I = \int_{-1}^1 \frac{\log (1+x)}{1+x^2} d x$.Split the integral at $x=0$:$I = \int_{-1}^0 \frac{\log (1+x)}{1+x^2} d x + \int_0^1 \frac{\log (1+x)}{1+x^2} d x$.Let $I_1 = \int_{-1}^0 \frac{\log (1+x)}{1+x^2} d x$.Substitute $x = -t$, so $d x = -d t$. When $x = -1, t = 1$ and when $x = 0, t = 0$.$I_1 = \int_1^0 \frac{\log (1-t)}{1+(-t)^2} (-d t) = \int_0^1 \frac{\log (1-t)}{1+t^2} d t$.By the property of definite integrals $\int_a^b f(t) d t = \int_a^b f(x) d x$, we have $I_1 = \int_0^1 \frac{\log (1-x)}{1+x^2} d x$.Comparing this with the given equation $\int_{-1}^1 \frac{\log (1+x)}{1+x^2} d x = \int_0^1 \frac{\log (1+x)}{1+x^2} d x + \int_0^1 f(x) d x$, we identify $f(x) = \frac{\log (1-x)}{1+x^2}$.

$\int_{-1}^1 \frac{\log (1+x)}{1+x^2} d x = \int_0^1 \frac{\log (1+x)}{1+x^2} d x + \int_0^1 f(x) d x$, then $f(x) =$

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