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Question Number 45373 by arvinddayama01@gmail.com last updated on 12/Oct/18

∫(t^3 /(1+t))dt=?

$$\int\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}=? \\ $$

Commented by maxmathsup by imad last updated on 12/Oct/18

∫ (t^3 /(1+t))dt =∫ ((t^3 +1−1)/(1+t))dt =∫ (t^2 −t+1)dt−∫  (dt/(1+t))  =(t^3 /3) −(t^2 /2)  +t −ln∣1+t∣ +k .

$$\int\:\frac{{t}^{\mathrm{3}} }{\mathrm{1}+{t}}{dt}\:=\int\:\frac{{t}^{\mathrm{3}} +\mathrm{1}−\mathrm{1}}{\mathrm{1}+{t}}{dt}\:=\int\:\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right){dt}−\int\:\:\frac{{dt}}{\mathrm{1}+{t}} \\ $$$$=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\:−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:\:+{t}\:−{ln}\mid\mathrm{1}+{t}\mid\:+{k}\:. \\ $$

Answered by ajfour last updated on 12/Oct/18

∫(((1+t)(1−t+t^2 ))/(1+t))dt−∫(dt/(1+t))  = t−(t^2 /2)+(t^3 /3)−ln ∣1+t∣+c .

$$\int\frac{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}−{t}+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}}{dt}−\int\frac{{dt}}{\mathrm{1}+{t}} \\ $$$$=\:{t}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{ln}\:\mid\mathrm{1}+{t}\mid+{c}\:. \\ $$

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