Question Number 93762 by ckkim89 last updated on 14/May/20
Answered by maths mind last updated on 14/May/20
$$=\frac{\mathrm{4}{xe}^{\mathrm{2}{x}} }{\mathrm{4}\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}.\frac{{de}^{\mathrm{2}{x}} .\left(\mathrm{1}+\mathrm{2}{x}\right)−{d}\left(\mathrm{1}+\mathrm{2}{x}\right).{e}^{\mathrm{2}{x}} }{\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\int\frac{{xe}^{\mathrm{2}{x}} }{\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{4}}\int{d}\left(\frac{{e}^{\mathrm{2}{x}} }{\mathrm{1}+\mathrm{2}{x}}\right)=\frac{\mathrm{1}}{\mathrm{4}}.\frac{{e}^{\mathrm{2}{x}} }{\mathrm{1}+\mathrm{2}{x}}+{c} \\ $$
Commented by ckkim89 last updated on 14/May/20
It's awsome!!! Thanks!!
Commented by M±th+et+s last updated on 15/May/20
$${nice}\:{to}\:{see}\:{you}\:{again}\:{sir}\: \\ $$
Commented by maths mind last updated on 18/May/20
$${i}\:{will}\:{back}\:{soon}\:{hope}\:{so}\: \\ $$$${i}/{worck}\:\:{verry}\:{hard}\:{these}\:{days}\:{too}\:{busy}\: \\ $$
Commented by M±th+et+s last updated on 18/May/20
$${god}\:{bless}\:{your}\:{work}\:{sir} \\ $$