int_{0}^{1} tan^{-1} left( frac{1}{x^2 - x + | vedclass

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(D) ધારો કે $I = \int_{0}^{1} 	an^{-1} \left( \frac{1}{x^2 - x + 1} ight) dx$. નિત્યસમ $	an^{-1} a - 	an^{-1} b = 	an^{-1} \left( \frac{a - b}{1

Vedclass · Accepted Answer

$\frac{\pi}{2} - \ln 2$

Vedclass · Answer

(D) ધારો કે $I = \int_{0}^{1} 	an^{-1} \left( \frac{1}{x^2 - x + 1} ight) dx$. નિત્યસમ $	an^{-1} a - 	an^{-1} b = 	an^{-1} \left( \frac{a - b}{1 + ab} ight)$ નો ઉપયોગ કરતા: $\frac{1}{x^2 - x + 1} = \frac{x - (x - 1)}{1 + x(x - 1)}$. તેથી,$	an^{-1} \left( \frac{1}{x^2 - x + 1} ight) = 	an^{-1} x - 	an^{-1} (x - 1)$. $I = \int_{0}^{1} 	an^{-1} x dx - \int_{0}^{1} 	an^{-1} (x - 1) dx$. બીજા સંકલન માટે ગુણધર્મ $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$ નો ઉપયોગ કરતા: $\int_{0}^{1} 	an^{-1} (x - 1) dx = \int_{0}^{1} 	an^{-1} ((1 - x) - 1) dx = \int_{0}^{1} 	an^{-1} (-x) dx = -\int_{0}^{1} 	an^{-1} x dx$. આ કિંમત $I$ માં મૂકતા: $I = \int_{0}^{1} 	an^{-1} x dx - (-\int_{0}^{1} 	an^{-1} x dx) = 2 \int_{0}^{1} 	an^{-1} x dx$. ખંડશઃ સંકલનનો ઉપયોગ કરતા: $\int 	an^{-1} x dx = x 	an^{-1} x - \frac{1}{2} \ln(1 + x^2)$. $I = 2 [x 	an^{-1} x - \frac{1}{2} \ln(1 + x^2)]_{0}^{1} = 2 [ (1 \cdot \frac{\pi}{4} - \frac{1}{2} \ln 2) - (0 - 0) ] = 2 [ \frac{\pi}{4} - \frac{1}{2} \ln 2 ] = \frac{\pi}{2} - \ln 2$.

$\int_{0}^{1} \tan^{-1} \left( \frac{1}{x^2 - x + 1} \right) dx$ ની કિંમત શોધો.

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