Question Number 37028 by nishant last updated on 08/Jun/18
$$\int\frac{\mathrm{2}{x}−\mathrm{1}}{\mathrm{5}{x}^{\mathrm{2}} −{x}+\mathrm{2}}\:{dx}\:\:=\:\:? \\ $$
Answered by ajfour last updated on 08/Jun/18
$${I}=\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\mathrm{10}{x}−\mathrm{1}}{\mathrm{5}{x}^{\mathrm{2}} −{x}+\mathrm{2}}{dx}\:−\frac{\mathrm{4}}{\mathrm{25}}\int\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{10}}\right)^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{100}}} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\:\mid\mathrm{5}{x}^{\mathrm{2}} −{x}+\mathrm{2}\mid−\frac{\mathrm{4}}{\mathrm{25}}\int\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{10}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{39}}}{\mathrm{10}}\right)^{\mathrm{2}} } \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\:\mid\mathrm{5}{x}^{\mathrm{2}} −{x}+\mathrm{2}\mid−\frac{\mathrm{4}}{\mathrm{25}}×\frac{\mathrm{10}}{\sqrt{\mathrm{39}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{10}{x}−\mathrm{1}}{\sqrt{\mathrm{39}}}\right)+{c}\:. \\ $$$$ \\ $$
Commented by nishant last updated on 08/Jun/18
$${thanks}\:{sir}. \\ $$