Evaluate-2y-4-y-3-y-2-y-1-dy-

Question Number 210308 by Spillover last updated on 05/Aug/24

$${Evaluate}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{\mathrm{2}{y}^{\mathrm{4}} }{{y}^{\mathrm{3}} −{y}^{\mathrm{2}} +{y}−\mathrm{1}}{dy} \\ $$

Commented by Frix last updated on 06/Aug/24

Simply expand it: =∫(2y+(1/(y−1))−(y/(y^2 +1))−(1/(y^2 +1))+2)dy= =y^2 +ln ∣y−1∣ −(1/2)ln (y^2 +1) −tan^(−1) y +2y+C= =y(y+2)+(1/2)ln (((y−1)^2 )/(y^2 +1)) −tan^(−1) y +C

$$\mathrm{Simply}\:\mathrm{expand}\:\mathrm{it}: \\ $$$$=\int\left(\mathrm{2}{y}+\frac{\mathrm{1}}{{y}−\mathrm{1}}−\frac{{y}}{{y}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{{y}^{\mathrm{2}} +\mathrm{1}}+\mathrm{2}\right){dy}= \\ $$$$={y}^{\mathrm{2}} +\mathrm{ln}\:\mid{y}−\mathrm{1}\mid\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({y}^{\mathrm{2}} +\mathrm{1}\right)\:−\mathrm{tan}^{−\mathrm{1}} \:{y}\:+\mathrm{2}{y}+{C}= \\ $$$$={y}\left({y}+\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{\left({y}−\mathrm{1}\right)^{\mathrm{2}} }{{y}^{\mathrm{2}} +\mathrm{1}}\:−\mathrm{tan}^{−\mathrm{1}} \:{y}\:+{C} \\ $$

Answered by Spillover last updated on 06/Aug/24