સ્તંભ I માં આપેલા સંકલિતોને સ્તંભ II માં આપેલ | vedclass

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(A) $(A) \int_{-1}^1 \frac{dx}{1+x^2} = [	an^{-1} x]_{-1}^1 = 	an^{-1}(1) - 	an^{-1}(-1) = \frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2}$. તેથી,

Vedclass · Accepted Answer

$A-s, B-s, C-p, D-r$

Vedclass · Answer

(A) $(A) \int_{-1}^1 \frac{dx}{1+x^2} = [	an^{-1} x]_{-1}^1 = 	an^{-1}(1) - 	an^{-1}(-1) = \frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2}$. તેથી,$A-s$.$(B) \int_0^1 \frac{dx}{\sqrt{1-x^2}} = [\sin^{-1} x]_0^1 = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$. તેથી,$B-s$.$(C) \int_2^3 \frac{dx}{1-x^2} = \int_2^3 \frac{dx}{-(x^2-1)} = -[\frac{1}{2} \ln |\frac{x-1}{x+1}|]_2^3 = -\frac{1}{2} [\ln(\frac{2}{4}) - \ln(\frac{1}{3})] = -\frac{1}{2} [\ln(\frac{1}{2}) - \ln(\frac{1}{3})] = -\frac{1}{2} \ln(\frac{1/2}{1/3}) = -\frac{1}{2} \ln(\frac{3}{2}) = \frac{1}{2} \ln(\frac{2}{3})$. તેથી,$C-p$.$(D) \int_1^2 \frac{dx}{x \sqrt{x^2-1}} = [\sec^{-1} x]_1^2 = \sec^{-1}(2) - \sec^{-1}(1) = \frac{\pi}{3} - 0 = \frac{\pi}{3}$. તેથી,$D-r$.

સ્તંભ $I$	સ્તંભ $II$
$(A) \int_{-1}^1 \frac{dx}{1+x^2}$	$(p) \frac{1}{2} \log \left(\frac{2}{3}\right)$
$(B) \int_0^1 \frac{dx}{\sqrt{1-x^2}}$	$(q) 2 \log \left(\frac{2}{3}\right)$
$(C) \int_2^3 \frac{dx}{1-x^2}$	$(r) \frac{\pi}{3}$
$(D) \int_1^2 \frac{dx}{x \sqrt{x^2-1}}$	$(s) \frac{\pi}{2}$

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જો $I_1 = \int_0^{\pi / 2} \frac{x}{\sin x} dx$ અને $I_2 = \int_0^1 \frac{\tan^{-1} x}{x} dx$ હોય,તો $I_1 : I_2$ શું થાય?

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