The value of the integral int_{0}^{1} x cot^{ | vedclass

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(A) Let $I = \int_{0}^{1} x \cot^{-1}(1 - x^2 + x^4) dx$.Using the property $\cot^{-1}(u) = 	an^{-1}(\frac{1}{u})$ for $u > 0$,we get:$I = \int_{0}^{

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$\frac{\pi}{4} - \frac{1}{2} \log_e 2$

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(A) Let $I = \int_{0}^{1} x \cot^{-1}(1 - x^2 + x^4) dx$.Using the property $\cot^{-1}(u) = 	an^{-1}(\frac{1}{u})$ for $u > 0$,we get:$I = \int_{0}^{1} x 	an^{-1}\left(\frac{1}{1 - x^2 + x^4}ight) dx$.We can write the argument as $\frac{x^2 - (x^2 - 1)}{1 + x^2(x^2 - 1)}$.Thus,$I = \int_{0}^{1} x \left[ 	an^{-1}(x^2) - 	an^{-1}(x^2 - 1) ight] dx$.Let $x^2 = t$,then $2x dx = dt$,so $x dx = \frac{1}{2} dt$.When $x=0, t=0$ and when $x=1, t=1$.$I = \frac{1}{2} \int_{0}^{1} 	an^{-1}(t) dt - \frac{1}{2} \int_{0}^{1} 	an^{-1}(t - 1) dt$.Using the property $\int_{0}^{a} f(t) dt = \int_{0}^{a} f(a - t) dt$ on the second integral:$I = \frac{1}{2} \int_{0}^{1} 	an^{-1}(t) dt - \frac{1}{2} \int_{0}^{1} 	an^{-1}(1 - t - 1) dt = \frac{1}{2} \int_{0}^{1} 	an^{-1}(t) dt - \frac{1}{2} \int_{0}^{1} 	an^{-1}(-t) dt$.Since $	an^{-1}(-t) = -	an^{-1}(t)$:$I = \frac{1}{2} \int_{0}^{1} 	an^{-1}(t) dt + \frac{1}{2} \int_{0}^{1} 	an^{-1}(t) dt = \int_{0}^{1} 	an^{-1}(t) dt$.Integrating by parts: $\int 	an^{-1}(t) dt = t 	an^{-1}(t) - \int \frac{t}{1+t^2} dt = t 	an^{-1}(t) - \frac{1}{2} \ln(1+t^2)$.Evaluating from $0$ to $1$:$I = [1 \cdot 	an^{-1}(1) - \frac{1}{2} \ln(2)] - [0 - 0] = \frac{\pi}{4} - \frac{1}{2} \ln 2$.

The value of the integral $\int_{0}^{1} x \cot^{-1}(1 - x^2 + x^4) dx$ is

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