Question Number 22210 by cmaxamuud98 @gmail.com last updated on 13/Oct/17
$$\int\frac{\mathrm{4}{x}}{{e}^{\mathrm{3}{x}} }{dx} \\ $$
Answered by ajfour last updated on 13/Oct/17
$$=\mathrm{4}\int{xe}^{−\mathrm{3}{x}} {dx} \\ $$$$=\mathrm{4}\left\{{x}\int{e}^{−\mathrm{3}{x}} {dx}−\int\left(\int{e}^{−\mathrm{3}{x}} {dx}\right){dx}\right\} \\ $$$$=\mathrm{4}\left\{{x}\left(−\frac{\mathrm{1}}{\mathrm{3}}{e}^{−\mathrm{3}{x}} +{c}_{\mathrm{1}} \right)−\int\left(\frac{{e}^{−\mathrm{3}{x}} }{−\mathrm{3}}+{c}_{\mathrm{1}} \right){dx}\right\} \\ $$$$=\mathrm{4}\left\{−\frac{{x}}{\mathrm{3}}{xe}^{−\mathrm{3}{x}} +{c}_{\mathrm{1}} {x}−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{{e}^{−\mathrm{3}{x}} }{−\mathrm{3}}+{c}_{\mathrm{1}} \right)\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{c}_{\mathrm{1}} {x}+{c}_{\mathrm{2}} \right\} \\ $$$$=−\frac{\mathrm{4}}{\mathrm{3}}\boldsymbol{{xe}}^{−\mathrm{3}\boldsymbol{{x}}} +\frac{\mathrm{4}}{\mathrm{9}}\boldsymbol{{e}}^{−\mathrm{3}\boldsymbol{{x}}} +\boldsymbol{{c}}\: \\ $$$$=\frac{\mathrm{4}}{\mathrm{9}}\boldsymbol{{e}}^{−\mathrm{3}\boldsymbol{{x}}} \left(\mathrm{1}−\mathrm{3}\boldsymbol{{x}}\right)+\boldsymbol{{c}}\:. \\ $$