Question Number 83250 by mathmax by abdo last updated on 29/Feb/20
$$\left.\mathrm{1}\right){decompose}\:{F}\left({x}\right)=\frac{{x}^{\mathrm{2}} −\mathrm{3}}{\mathrm{2}{x}^{\mathrm{3}} \:+\mathrm{5}{x}+\mathrm{7}} \\ $$$$\left.\mathrm{2}\right){determine}\:\int\:{F}\left({x}\right){dx} \\ $$
Answered by MJS last updated on 29/Feb/20
$$\frac{{x}^{\mathrm{2}} −\mathrm{3}}{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}+\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{11}}\left(\frac{\mathrm{15}{x}−\mathrm{19}}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{7}}−\frac{\mathrm{2}}{{x}+\mathrm{1}}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{11}}\left(\frac{\mathrm{15}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\mathrm{4}\left({x}^{\mathrm{2}} −{x}+\frac{\mathrm{7}}{\mathrm{2}}\right)}−\frac{\mathrm{23}}{\mathrm{4}\left({x}^{\mathrm{2}} −{x}+\frac{\mathrm{7}}{\mathrm{2}}\right)}−\frac{\mathrm{2}}{{x}+\mathrm{1}}\right) \\ $$$$\int\frac{{x}^{\mathrm{2}} −\mathrm{3}}{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}+\mathrm{7}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{11}}\left(\frac{\mathrm{15}}{\mathrm{4}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\frac{\mathrm{7}}{\mathrm{2}}\right)\:−\frac{\mathrm{23}}{\mathrm{2}\sqrt{\mathrm{13}}}\mathrm{arctan}\:\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{13}}}\:−\mathrm{2ln}\:\left({x}+\mathrm{1}\right)\right)= \\ $$$$=\frac{\mathrm{15}}{\mathrm{44}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −{x}+\frac{\mathrm{7}}{\mathrm{2}}\right)\:−\frac{\mathrm{2}}{\mathrm{11}}\mathrm{ln}\:\mid{x}+\mathrm{1}\mid\:−\frac{\mathrm{23}\sqrt{\mathrm{13}}}{\mathrm{286}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{13}}\left(\mathrm{2}{x}−\mathrm{1}\right)}{\mathrm{13}}\:+{C} \\ $$
Commented by mathmax by abdo last updated on 29/Feb/20
$${thanks}\:{sir}\:{mjs} \\ $$