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Category: Integration

dx-x-2-1-x-4-1-

Question Number 133872 by MJS_new last updated on 24/Feb/21 $$\int\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}=? \\ $$ Commented by MJS_new last updated on 24/Feb/21 $$\mathrm{I}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{but}\:\mathrm{maybe}\:\mathrm{there}'\mathrm{s}\:\mathrm{an}\:\mathrm{easier}\:\mathrm{path}… \\ $$ Commented…

4sin-3x-e-4x-4-

Question Number 68316 by 9102176137086 last updated on 08/Sep/19 $$\int\left(\mathrm{4sin}\:\mathrm{3}{x}+\frac{{e}^{\mathrm{4}{x}} }{\mathrm{4}}\right) \\ $$ Commented by mathmax by abdo last updated on 08/Sep/19 $$=\mathrm{4}\int\:{sin}\left(\mathrm{3}{x}\right){dx}\:+\frac{\mathrm{1}}{\mathrm{4}}\int\:{e}^{\mathrm{4}{x}} {dx}\:+{c} \\…

1-6-x-2-x-2-x-

Question Number 68313 by 9102176137086 last updated on 08/Sep/19 $$\int\left(\mathrm{1}−\frac{\mathrm{6}}{{x}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }+\sqrt{{x}}\right) \\ $$ Commented by mathmax by abdo last updated on 10/Sep/19 $$\int\:\left(\mathrm{1}−\frac{\mathrm{6}}{{x}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\:+\sqrt{{x}}\right){dx}\:={x}−\mathrm{6}{ln}\mid{x}\mid\:+\frac{\mathrm{2}}{\mathrm{3}}{x}^{\frac{\mathrm{3}}{\mathrm{2}\:}} \:+{c}…

advanced-integral-prove-that-0-1-e-x-1-e-x-dx-x-Golden-ratio-

Question Number 133791 by mnjuly1970 last updated on 24/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:……{advanced}\:\:\:\:{integral}…. \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\mathrm{1}−{e}^{−\varphi{x}} }{\mathrm{1}+{e}^{\varphi{x}} }\:\right)\frac{{dx}}{{x}}\:=?? \\ $$$$\:\:\:\varphi:\:=\:{Golden}\:{ratio}… \\ $$$$ \\ $$ Answered…

A-0-1-sin-1-x-2-1-2x-4-2-dx-

Question Number 133786 by bobhans last updated on 24/Feb/21 $$\mathcal{A}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\:\sqrt{\mathrm{2}{x}^{\mathrm{4}} +\mathrm{2}}}\:\right)\:{dx}\:=? \\ $$ Answered by john_santu last updated on 24/Feb/21 $${Using}\:{the}\:{Pythagorean}\:{theorem}\:…