Question Number 120064 by bemath last updated on 29/Oct/20
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Question Number 120060 by bemath last updated on 29/Oct/20
Question Number 120059 by A8;15: last updated on 29/Oct/20 Answered by Lordose last updated on 29/Oct/20
Question Number 120040 by bramlexs22 last updated on 28/Oct/20
Question Number 120025 by mathmax by abdo last updated on 28/Oct/20
Question Number 119989 by mnjuly1970 last updated on 28/Oct/20 Answered by mathmax by abdo last updated on 28/Oct/20 $$\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{xdx}}{\mathrm{2e}^{\mathrm{x}} −\mathrm{1}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{xe}^{−\mathrm{x}} \mathrm{dx}}{\mathrm{2}−\mathrm{e}^{−\mathrm{x}}…
Question Number 119970 by bramlexs22 last updated on 28/Oct/20
Question Number 119960 by mnjuly1970 last updated on 28/Oct/20 Answered by mathmax by abdo last updated on 28/Oct/20 $$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctanx}}{\left(\mathrm{1}+\mathrm{x}\right)\sqrt{\mathrm{x}}}\mathrm{dx}\:\Rightarrow\mathrm{A}\:=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}}…
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