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Question Number 71091 by Henri Boucatchou last updated on 11/Oct/19

Commented by Prithwish sen last updated on 11/Oct/19

It is same as  ∫_0 ^∞ (1/x)tan^(−1) (√(x^2 +x+1)) dx −∫_0 ^∞ (1/x)tan^(−1) (√(9x^2 +3x+1)) dx

$$\boldsymbol{\mathrm{It}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{same}}\:\boldsymbol{\mathrm{as}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}}\:\boldsymbol{\mathrm{dx}}\:−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\boldsymbol{\mathrm{tan}}^{−\mathrm{1}} \sqrt{\mathrm{9}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{1}}\:\boldsymbol{\mathrm{dx}} \\ $$

Commented by mind is power last updated on 11/Oct/19

yes nice

$$\mathrm{yes}\:\mathrm{nice}\: \\ $$$$ \\ $$

Commented by Prithwish sen last updated on 11/Oct/19

Thank you Sir.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}. \\ $$

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