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Category: Integration

dx-x-1-3-x-2-5-1-4-

Question Number 159315 by cortano last updated on 15/Nov/21 $$\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{4}}]{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}+\mathrm{2}\right)^{\mathrm{5}} }\:}\:? \\ $$ Commented by tounghoungko last updated on 15/Nov/21 $${Y}=\int\:\frac{{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}\right)^{\mathrm{3}/\mathrm{4}} }\: \\…

Question-93762

Question Number 93762 by ckkim89 last updated on 14/May/20 Answered by maths mind last updated on 14/May/20 $$=\frac{\mathrm{4}{xe}^{\mathrm{2}{x}} }{\mathrm{4}\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}.\frac{{de}^{\mathrm{2}{x}} .\left(\mathrm{1}+\mathrm{2}{x}\right)−{d}\left(\mathrm{1}+\mathrm{2}{x}\right).{e}^{\mathrm{2}{x}} }{\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\int\frac{{xe}^{\mathrm{2}{x}}…

Question-159297

Question Number 159297 by rs4089 last updated on 15/Nov/21 Commented by rs4089 last updated on 15/Nov/21 $${how}\:{can}\:{i}\:{find}\:{slope}\:{and}\:{deflection} \\ $$$$\:{of}\:{this}\:{cantilever}\:{beam}\:{at}\:{free}\:{end}\: \\ $$$${point}.\:{by}\:{using}\:{double}\:{integral}\:{method} \\ $$ Answered by…

let-give-I-0-1-ln-1-x-1-x-2-dx-and-J-0-1-2-x-1-x-2-1-xy-dxdy-calculate-J-by-two-methods-then-find-the-value-of-I-

Question Number 28200 by abdo imad last updated on 21/Jan/18 $${let}\:{give}\:{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{and}\:{J}=\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\frac{{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{xy}\right)}{dxdy} \\ $$$${calculate}\:{J}\:{by}\:{two}\:{methods}\:{then}\:{find}\:{the}\:{value}\:{of}\:{I}. \\ $$ Commented by abdo imad…

dx-x-1-3-x-2-2x-

Question Number 93720 by john santu last updated on 14/May/20 $$\int\:\frac{\boldsymbol{{dx}}}{\left(\boldsymbol{{x}}+\mathrm{1}\right)^{\mathrm{3}} \:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}}}\: \\ $$ Commented by john santu last updated on 14/May/20 $$\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}\:=\:\boldsymbol{{t}}^{\mathrm{2}\:}…

Question-159233

Question Number 159233 by mnjuly1970 last updated on 14/Nov/21 Answered by mindispower last updated on 16/Nov/21 $${A}_{\mathrm{2}{n}} =\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}\right)^{\mathrm{2}{n}} }=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} }.\zeta\left(\mathrm{2}{n}\right) \\ $$$$\frac{\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}} }}{\mathrm{2}{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}=−\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)}+\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}}…