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Miscellaneous Exercise 4 · Q82

Q.If ∫0k12+8x2 dx=π16\int_0^k \dfrac{1}{2+8x^2}\,dx = \dfrac{\pi}{16}, find kk.

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∫0kdx2+8x2=∫0kdx2(1+4x2)=14[tan⁡−1(2x)]0k=14tan⁡−1(2k).\int_0^k\frac{dx}{2+8x^2}=\int_0^k\frac{dx}{2(1+4x^2)}=\frac14\big[\tan^{-1}(2x)\big]_0^k=\frac14\tan^{-1}(2k). …

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